Pascack Valley Regional School District
Bilash - Burns
Midterm Examination 2010 Outline & Instructions
The midterm exam will consist of two components:
Part 1] 50 multiple choice questions that include short answer and short problems requiring calculations.
This section covers all topics from chapter 1-5 (Giancoli) -- covered in class. Each question is worth 1
point.
Part 2] Complete 5 of 6 free response problems that will be taken from the following ten topics:
Horizontal Kinematics
Free Fall Kinematics
Kinematics of Projectile Launched Horizontally
Kinematics of Projectiles Launched at Angles
Vector’s & Newton’s Laws (Static Equilibrium)
Newton’s Laws: Atwood’s Machines
Newton’s Laws & Incline Planes
Newton’s Laws & Friction: An Experiment to be Designed
Circular Motion
Newton’s Law of Gravity
Each problem must be solved in detail including the derivation of applicable formulas that are not listed in
the formula sheet included below. Each problem is worth I0 points.
The following formulas will be supplied on the exan~
v~=vo2 + 2ax g = -9.8 m/s~
v = vo + at x=½ (v + vo)t x=vot + ½at2
2sinOcosO = sin20 R = vo2 sin20/g
l fl = 0.3048 m
=ktN
ZF = ma
w=mg
FU
FAD = -FBA
C = 2~r
a~ = vr2/r F~ = mac
Vr = C/T
2
~
F~ = Gmzm2/r G = 6,67 x lO4Z N.m~/kg re = 6380 km me = 5.98 x l O24 kg
v=Ax/At
Please bring the items listed below to the exam. No other items will be permitted.
¯ 1 calculator, with fresh batteries and/or extra batteries
¯ Three #2 pencils, with erasers
¯ WhiteOut, to cover erasure marks on the Scantron sheets (don’t rely on erasers[)
[CHAP. 4
UNIFORMLY ACCELERATED MOTION
32
INSTANTANEOUS VELOCITY is the average velocity evaluated for a time interval that
approaches zero. Thus, if an object undergoes a displacement Ax in a time At, then for that object,
Ax
v = instantaneous velocity = tim -~t-~O At
where the notation means that the ratio Ax/At is to be evaluated for a time interval At that
approaches zero.
GRAPHICAL INTERPRETATIONS for motion along a straight line (the x-axis) are as follows:
The instantaneous velocity of an object at a certain time is the slope of the x-versus-t graph at
that time.
The instantaneous acceleration of an object at a certain time is the slope of the v-versus-t graph at
that time.
For constant-velocity motion, the x-versus-t graph is a straight line. For constant-acceleration
motion, the v-versus-t graph is a straight line.
ACCELERATION DUE TO GRAVITY (g): The acceleration of a body moving only under the
force of gravity is g, the gravitational (or free-fall) acceleration, which is directed verticall
downward. On earth, g = 9.8 m/sz (= 32.2ft/sa); the value varies slightly from place to place. On
the moon, the free-fall acceleration is 1.6 m/sa.
PROJECTILE PROBLEMS can be solved easily if air friction can be ignored. One simply considers
the motion to consist of two independent parts: horizontal motion with a = 0 and vr = v0 = ~ (i.e.,
constant velocity), and vertical motion with a =g = 9.8 m/sz downward.
Solved Problems
4.1
Change the speed 0.200 cm/s to units of km/year.
year
4°2
A runner makes one lap around a 200 m track in a time of 25 s. What were the runner’s (a)
average speed and (b) average velocity?
(a) From the definition,
distance traveled 200 m
= 8.00 m/s
Average speed = time taken = -25 s
(b)
Because the run ended at the starting point, the displacement vector from starting point to end
point has zero length. Therefore,
Lisplacement = 0m = 0 m/s
time 25 s
UNIFORMLY ACCELERATED MOTION
CHAP. 4]
33
An object starts from rest with constant acceleration 8 m/sz along a straight line. Find (a) the
speed at the end of 5 s, (b) the average speed for the 5 s interval, and (c) the distance traveled
in the 5 s.
We are interested in the motion for the first 5s. Take the direction of motion to be the
+x-direction. We know that Vo = 0, t = 5 s, and a = 8 m/s2. Because the motion is uniformly accelerated,
the five motion equations apply.
(a)
v~. = Vo + at = 0 + (8 m/s2)(5 s) = 40 m/s
(b)
~ = vo + vr = 0 + 40 m/s = 20 m/s
2
2
x = rot + ½at~ = 0 + ~(8 m/s~)(5 s)z = 100 m or x = ~’t = (20 m/s)(5 s) = 100 m
(c)
4,4
A truck’s speed increases uniformly from 15 km/h to 60 km/h in 20 s. Determine (a) the
average speed, (b) the acceleration, and (c) the distance traveled, al! in units of meters and
seconds.
For the 20 s trip under discussion, taking +x to be in the direction of motion, we have
v0 =.(15 ~)(1000 ,~rfi)(3~00 ~’~)= 4.17 m/s
v~= 60 km/h = 16.7 m/s
t=20s
(a)
(b)
(c)
~ = ½(vo + v~) = ½(4.17 + 16.7) m/s = 10.4 m/s
a vz - vo (16.7-4.2) m/s= 0.63 m/s2
=
=
20 s
x = Ot = (10.4 m/s)(20 s) = 208 m
The graph of an object’s motion along a line is shown in Fig. 4-1. Find the instantaneous
velocity of the object at points A and B. What is the object’s average velocity? Its
acceleration?
lax=4 ra
0
Because the velocity is given by the slope Ax]At of the tangent line, we take a tangent to the curve
at point A. The tangent line is the curve itself in this case. For the triangle shown at A, we have
Ax 4m
.... 0.50 m/s
At 8s
This is also the velocity at point B and at every other point on the straight-line graph. It follows that
a = 0 and ~ = v~ = 0.50 m/s.
[CHAP. 4
UNIFORMLY ACCELERATED MOTION
34
~An
object’s motion along the x-axis is graphed in Fig. 4-2. Describe its motion.
The velocity of the object at any instant is equal to the slope of the graph at the point corresponding
to that instant. Because the slope is zero from t = 0 s to t = 2 s, the object is standing still during this
time interval. At t = 2 s, the object begins to move in the +x-direction with constant velocity (the slope
is positive and constant). For the interval t = 2 s to t = 4 s,
-v=slope
......... rise x~--x0 3,0m-0m 3.0m1.50 m/s
run q-to 4.0s-2.0s 2.0s
During the interval t = 4 s to t = 6 s, the object is at rest; the slope of the graph is zero and x does
~ change for that interval.
From t = 6 s to t = 10 s and beyond, the object is moving in the -x-direction; the slope and the
velocity are negative. We have
v
- =slope
........ x~-Xo -2,0m-3.0m -5,0m 1.25 m/s
t~ - to 10.0 s - 6.0 s 4.0 s
4
8
6
4
2
-2
!0
5
15
Fig. 4-3
Fig. 4-2
The vertical motion of an object is graphed in Fig. 4-3. Describe its motion qualitatively, and
find its instantaneous velocity at points A, B, and C.
Recalling that the instantaneous velocity is given by the slope of the graph, we see that the object is
moving fastest at t = 0. As it rises, it slows and finally stops at B. (The slope there is zero.) Then it
begins to fall back downward at ever-increasing speed.
At point A, we have
......... 2.25Ay
m/s 12.0m-3.0m 9.0m
VA-- slope-- At- 4.0s-0s
4,0s
The velocity at A is positive, so it is in the +y-direction. At points B and C,
slope = 0 m/s
Ay 5.5 m- 13.0 m = -7.5 m = _1.15 m/s
vc= sl°pe = A~= 15.0s- 8.5 s 6.5s
Because it is negative, the velocity at C is in the -y-direction.
A ball is dropped from rest at a height of 50m above the ground. (a) What is its speed just
before it hits the ground? (b) How long does it take to reach the ground?
UNIFORMLY ACCELERATED MOTION
CHAP. 4]
35
If we can ignore air friction, the ball is uniformly accelerated until it reaches the ground. Its
acceleration is downward and is 9.8 m/s2. Taking down as positive, we have for the trip:
y =50m a = 9.8m/s~ v0=0
v~ = v~ + 2ay = 0 + 2(9.8 m/s~)(50 m) = 980 m2/s2
(a)
and so v~ = 31.3 m/s.
(b) From a = (vf - vo)/t,
t = v! - v~o = (31.3 - 0) m/s = 3.19 s
a
9.8 m/s2
(We could just as well have taken up as positive. How would the caiculafion have been changed?)
A skier starts from rest and slides 9 m down a slope in 3 s. In what time after starting will the
skier acquire a velocity of 24 m/s? Assume constant acceleration.
We must find the skier’s acceleration from the data concerning the 3 s trip. Taking the direction of
motion as the +x-direction, we have t = 3 s, vo = 0, and x = 9 m, Then x = rot + ½at2 gives
2x 18m
,2
a = ~ = (~s)~ = 2.00 m/s
We can now use this value of a for the longer trip, from the starting point to the place where
v = 24 m/s. For this trip, Vo = 0, vy = 24 m/s, a = 2 m/sz. Then, from v¢ = vo + at,
- 242m/s_
t -- v.~
~ --2vom/s
,- 12.0 s
(~(/4/.~0~ A bus moving at a speed of 20 m/s begins to slow at a rate of 3 m/s each second. Find how far
:~~\j j it goes before stopping.
Take the direction of motion to be the +x-direction. For the trip under consideration, vo = 20 m/s,
W = 0 m/s, a = -3 m/sz. Notice that the bus is not speeding up in the positive motion direction. Instead,
it is slowing in that direction and so its acceleration is negative (a deceleration). Use
to find
-(20 m/s)~ = 66.7 m
x = 2(-3 m/s2)
A car moving at 30 m/s slows uniformly to a speed of 10 m/s in a time of 5s. Determine (a)
the acceleration of the car and (b) the distance it moves in the third second.
Let us take the direction of motion to be the +x direction
(a)
Forthe5sinterval, wehavet=5s, v~=3Om/s,v~=lOm/s. Usingv¢=vo+atgives
a = (10 - 30) m/s _ -4.00 m/s~
5s
(b)
x = (distance covered in 3 s) - (distance covered in 2 s)
= (rot3 + ½ate) - (rot2 + ½ate)
= Oo(t3 - tz) + ½a(t~ - t~)
Using Vo = 30 m/s, a = -4 m/sz, tz = 2 s, t~ = 3 s gives
x = (30 m/s)(l s) - (2 m/s~)(5 s2) = 2.00 m
The velocity of a train is reduced uniformly from 15 m/s to 7 m/s while traveling a distance of
UNIFORMLY ACCELERATED MOTION
36
[CHAP. 4
90 m. (a) Compute the acceleration. (b) How much farther will the train travel before coming
to rest, provided the acceleration remains constant?
Let us take the direction of motion to be the +x-direction.
(a) We have vo = 15 m/s, vs = 7 m/s, x = 90 m. Then @ = v0~ + 2ax gives
a = -0.98 m/s~
(b) We now have the new conditions vo = 7 m]s, vs = 0, a = -0.98 m/s2. Then
v~ = v~ + 2ax
gives
2
x - 0-(Tm/s)~ 25m
(/,~..13 ’I)A stone is thrown straight upward and it rises to a height of 20m. With what speed was it
throw°?
Let us take up as the positive y-direction. The stone’s velocity is zero at the top of its path~ Then
v~ = 0, y = 20 m, a = -9.8 m/sz. (The minos sign arises because the acceleration due to gravity is always
downward and we have taken up to be positive.)We use v} = v~ + 2ay to find
vo = ~/-2(-9.8 m/s2)(20 m) = 19.8 m/s
tl/4.14) A stone is thrown straight upward with a speed of 20m/s. It is caught on its way down at a
point 5.0 m above where it was thrown. (a) How fast was it going when it was caught? (b)
How long did the trip take?
The situation is shown in Fig. 4-4. Let us take up as positive. Then, for the trip that lasts from the
instant after throwing to the instant before catching, vo=20m/s, Y = +5m (since it is an upward
displacement), a = -9.8 m/s~.
(a) We use v}=v~+2ay to find
v~ = (20 m/s)2 + 2(-9.8 m/sZ)(5 m) = 302 mZ]sz
vs= :k~= -17.4 m/s
We take the negative sign because the stone is moving downward, in the negative direction, at the
final instant.
(b) We use a = (vs. - vo)lt to find
t = _(-17.4 - 20)~ m/s = 3.8 s
-9.8 m/s
Notice that we retain the minus sign on vs.
I Caught
I here
5ra
20ra 0~Fig. 4-4
CHAP. 4]
UNIFORMLY ACCELERATED MOTION
37
4o15,~J A ball that is thrown vertically upward on the moon returns to its starting point in 4 s. The
~~.~./~ acceleration due to gravity there is 1.60m/s~ downward. Find the bali’s original speed.
Let us take up as positive. For the trip from beginning to end, y = 0 (it ends at the same level it
started at), a = -1.60 m/s~, t = 4 s. We use y = rot + ½at~ to find
0 = vo(4 s) + ½(-1.60 m/sZ)(4 s)z
from which v0 = 3.20 m/s.
!A baseball is thrown straight upward on the moon with an initial speed of 35 m/s. Compute
(a) the maximum height reached by the ball, (b) the time taken to reach that height, (c) its
velocity 30 s after it is thrown, and (d) when the bali’s height is 100 m.
Take up as positive. At the highest point, the bali’s velocity is zero.
(a) From v~ = v~ + 2ay we have, since g = 1.6 m/sz on the moon,
0 = (35 m/s)~ + 2(-1.6 m/s2)y or
y = 383 m
(b) From v.~ = Vo + at we have
0 = 35 m/s + (-1.6 m/s2)t
or
t = 21.9 s
or
v; = -13.0 m/s
(c) From v.~ = vo + at we have
v:- = 35 m!s + (-1.6 m/s~)(30 s)
Because v,, is negative and we are taking up as positive, the velocity is directed downward. The ball
is on its way down at t = 30 s.
(d) From y = rot + ½at~ we have
100 m = (35 m/s)t + ½(- 1.6 m/sZ)tz or 0.80t2 - 35t + 100 = 0
By use of the quadratic formula,
x=
2a
we find t = 3.1 s and 40.6 s. At t = 3.1 s the ball is at 100 m and ascending; at t = 40.6 s it is at the
same height but descending.
A ballast bag is dropped from a balloon that is 300 m above the ground and rising at 13 m/s.
For the bag, find (a) the maximum height reached, (b) its position and velocity 5 s after it is
released, and (c) the time at which it hits the ground.
The initial velocity of the bag when released is the same as that of the balloon, 13 m/s upward. Let
us choose up as positive and take y = 0 at the point of release¯
(a) At the highest point, v; = 0. From v~ = v~ + 2ay,
O=(13m/s)2+2(-9.Sm/sZ)y
or y =8.6m
The maximum height is 300 + 8.6 = 308.6 m.
(b) Tilke the end point to be its position at t = 5 s. Then, from y = vot + ½atz,
y = (13 m/s)(5 s) + ½(-9.8 m/sZ)(5 s)2 = -57.5 m
So its height is 300 - 58 = 242 m. Also, from v~ = vo + at,
v~ = 13 m/s + (-9.8 m/s2)(5 s) = -36 m/s
It is on its way down with a velocity of 36 m/s downward.
(c) Just as it hits the ground, the bag’s displacement is -300 m. Then
y = rot + ½atz becomes -300 m = (13 m/s)t + ½(-9.8 m/s~)t2
UNIFORMLY ACCELERATED MOTION
38
[CHAP. 4
or 4.9tz- 13t-300 = 0. The quadratic formula gives t = 9.3 s and -6.6 s. Only the positive time
has physical meaning, so the required answer is 9.3 s.
We could have avoided the quadratic formula by first computing v,~:
v} = v~ + 2as becomes v~ = (13 m/s)~ + 2(-9.8 m/sZ)(-300 m)
so that v; = :1:77.8 m/s. Then, using the negative value for vj- (why?) in v,~ = vo + at gives t = 9.3 s,
as before.
As shown in Fig. 4-5, a projectile is fired with a horizontal velocity of 30 m/s from the top of a
cliff 80 m high. (a) How long will it take to strike the level ground at the base of the cliff? (b)
How far from the foot of the cliff will it strike? (~) With what veloci[~y will it strike?
I’o= 30 m/s
Fig. 4-5
(a) The horizontal and vertical motions are independent of each other. Consider first the vertical
motion. Taking up as positive, we have
or
-80 m = 0 + ~(-9.8 m/s2)tz
from which t = 4.04 s. Notice that the initial velocity had zero vertical component and so vo = 0 for
the vertical motion.
Now consider the horizontal motion. For it, a = 0 and so f,~ = v~ = vt~ = 30 m]s. Then, using the
value of t found in (a), we have
x = ~.t = (30 m/s)(4.04 s) = 121 m
The final velocity has a horizoutal component of 30 m/s. But its vertical component at t = 4.04 s is
given by v~ = roy + ayt as
vry = 0 + ( -9.8 m/sZ)(4.04 s) = -40 m!s
The resultant of these two components is labeled v in Fig. 4-5; we have
v = V(40 m/s)~ + (30 m/s)~ = 50.0 m/s
Angle 0 as shown is given by tan 0 = 40/30 and is 53°.
4.19 ~)A stunt flier is moving at 15 m/s parallel to the flat ground 100 m below, as Shown in Fig. 4-6.
~) How large must the distance x from plane to target be if a sack of flour released from the
~’~’~¢ plane is to strike the target?
Following the same procedure as in Problem 4.18, we use y = vo/+ ~a/z to get
-lfi0 m = 0 + ½(-9.8 m/sZ)fl"
Now )c = 9,t = (15 m/s)(4.52 s) = 67.8 ra.
or
t = 4.52 s
UNIFORMLY ACCELERATED MOTION
CHAP. 4]
Sack
39
15 m/s
Target
100 m
Fig. 4-7
Fig. 4-6
~A baseball is thrown with an initial velocity of 100 m/s at an angle of 30 above the horizontal,
as shown in Fig. 4-7. How far from the throwing point will the base,ball attain its original
level?
We divide the problem into horizontal and vertical parts, for which
Vox = Vo cos 30° = 86.6 m/s and roy = vo sin 30° = 50 m/s
where up is being taken as positive.
In the vertical problem, y = 0 since the ball returns to its original height. Then
! t2
y = Volt + za~
or 0 = (50 m/s) + ½(-9.8 m/s~)t
and t = 10.2 s.
In the horizontal problem, v0., =v~, = ~, = 86.6 m/s. Therefore,
x = ~t = (86.6 m/s)(10.2 s) = 884 m
As shown in Fig. 4-8, a ball is thrown from the top of one building toward a tail building 50 m
away. The initial velocity of the ball is 20 m!s at 40° above the horizontal. How ~ar above or
below its original level will the ball strike the opposite wall?
Fig. 4-8
We have
Vo. = (20 m/s) cos 40° = 15.3 m/s
Vo~. = (20 m/s) sin 400 = 12.9 m/s
Consider first the horizontal motion. For it,
vo~ =v~x = ~, = 15.3 m/s
Then x = 9,t gives
50 m = (15.3 m/s)t or
t =3.27s
For the vertical motion, taking down as positive, we have
y = volt + ~a~.t~ = (-!2.9 m/s)(3.27 s) + ½(9.8 m/s2)(3.27 s)2 = 105 m
Since y is positive, and since down is positive, the ball will hit at 105 m below the original level.
UNIFORMLY ACCELERATED MOTION
40
[CHAP. 4
!~(a) Find the range x of a gun which fires a shell with muzzle velocity v at an angle of elevation
0, (b) Find the angle of elevation 0 of a gun which fires a shell with a muzzle velocity of
120 m!s and hits a target on the same level but 1300 m distant. (See Fig. 4-9.)
Fig. 4-9
(a) Let t be the time it takes the shell to hit the target. Then, x = vast or t = X/Vox. Consider the
vertical motion alone, and take up as positive. When the shell strikes the target,
Vertical displacement = 0 = vort + ½(-g)t~
Solving this equation gives t = 2voy/g. But t = x/vox, so
2(vocos 0)(rosin 0)
x~=2Voy or
x =2v°~v~
g
g
vo~ g
The formula 2 sin 0 cos 0 = sin 20 can be used to simplify this. After substitution, we get
v~ sin 20
The maximum range corresponds to 0 = 45°; since sin 20 has a maximum value of 1 when 20 = 90°
or 0=45°¯
in (a), we have
(b) From the range equation found
slnzt,=~o~=
(120m/s)~-- 0.885
. ~., gx (9.8m/s~)(1300m)
Therefore, 20 = arcsin 0.885 = 62° and so 0 = 31°.
Supplementary Problems
4.23
A car’s odometer reads 22 687 km at the start of a trip and 22 791 km at the end. The trip took 4 hours.
What was the car’s average speed in km/h? In m/s? Ans. 26 km/h, 7.2 m/s
4.~
An auto travels at the rate of 25 km/h for 4 minutes, then at 50km/h for 8 minutes, and finally at
20 km/h for 2 minutes. Find (a) the total distance covered in km and (b) the average speed for the
complete trip in m/s. Ans. (a) 9 kin; (b) 10.7 m/s
4.25
A runner travels 1.5 laps around a circular track in a time of 50 s. The diameter of the track is 40 m, and
its circumference is 126 m. Find (a) the average speed of the runner and (b) the magnitude of the
runner’s average velodty. Ans. (a) 3.78 m/s; (b) 0.80 m/s
The following data describe the position of an object along the x-axis as a function of time. Plot the
data, and find the instantaneous velocity of the object at (a) t=5.0s, (b) 16.0s, and (c)
23.0 s. Ans. (a) 0.018 m/s; (b) 0 m/s; (c) -0.013 m/s
0 2 4 6 8 10 12 14 16 18 20 22 24 26 28
t, s
x, cm
0 4.0 7.8 11.3 14.3 16.8 18.6 19.7 20.0 19.5 18.2 16.2 13.5 10.3 6.7
4
CHAP. 4]
41
4.27
For the object whose motion is described in Problem 4.26, find its velocity at the following times: (a)
3.0s, (b) 10.0s, and (c) 24.0s Arts. (a) 1.9cm/s; (b) 1.1cm/s; (c) -1.5cm/s
4.28
For the object whose motion is plotted in Fig. 4-3, find its instantaneous velocity at the following times:
(a) t.0 s, (b) 4.0 s, and (c) 10.0 s. Arts. (a) 3.3 m/s; (b) 1.0 m/s; (c) -0.83 m/s
4.29
A body with initial velocity 8 m/s moves along a straight line with constant acceleration and travels
640m in 40s. For the 40s interval, find (a) the average velocity, (b) the final velocity, and (c) the
acceleration. Ans. (a) 16 m/s; (b) 24 m/s; (c) 0.40 m]sz
A truck starts from rest and moves with a constant acceleration of 5 m/sz. Find its speed and the
distance traveled after 4 s has elapsed. Ans. 20 m/s, 40 m
4.30
4.31
he
UNIFORMLY ACCELERATED MOTION
4.32
A box slides down an incline with uniform acceleration. It starts from rest and attains a speed of 2.7 m/s
in 3 s. Find (a) the acceleration and (b) the distance moved in the first 6 s.
Ans. (a) 0.90m/s2; (b) 16.2m
A car is accelerating uniformly as it passes two checkpoints that are 30 m apart. The time taken between \\~
checkpoints is 4.0 s, and the car’s speed at the first checkpoint is 5.0 m/s. Find the car’s acceleration and
its speed at the second checkpoint. Ans. 1.25 m/s~, 10 m/s
4.33
An auto’s velocity increases uniformly from 6.0 m/s to 20 m/s while covering 70 m. Find the acceleration
and the time taken. Ans. 2.6 m/sz, 5.4 s
4.34
A plane starts from rest and accelerates along the ground before takeoff. It moves 600 m in 12 s. Find
(a) the acceleration, (b) speed at the end of 12s, and (c) the distance moved during the twelfth
second. Ans. (a) 8.3 m/s2; (b) 100m/s; (c) 96m
4.35
A train running at 30 m/s is slowed uniformly to a stop in 44 s. Find the acceleration and the stopping
distance. Arts. -0.68 m/s2, 660 m
4.36
An object moving at 13m/s slows uniformly at the rate of 2m/s each second for a time of 6s.
Determine (a) its final speed, (b) its average speed during the 6 s, and (c) the distance moved in the
6s. Arts. (a) 1m/s; (b ) 7 m]s; (c) 42 m
4.37
A body falls freely from rest. Find (a) its acceleration, (b) the distance it falls in 3 s, (c) its speed after
falling 70m, (d) the time required to reach a speed of 25m/s, and (e) the time taken to fall
300 m. Arts. (a) 9.8 m/sZ; (b) 44 m; (c) 37 m/s; (d) 2.55 s; (e) 7.8 s
4.38
A marble dropped from a bridge strikes the water in 5 s. Calculate (a) the speed with which it strikes
and (b) the height of the bridge. Arts. (a) 49 m/s; (b) 123 m
4.39
A stone is thrown straight downward with initial speed 8 m/s from a height of 25 m. Find (a) the time it
takes to reach the ground and (b) the speed with which it strikes. Arts. (a) L59 s; (b) 23.5 m/s
4.40
A baseball is thrown straight upward with a speed of 30 m/s. (a) How long will it rise? (b) How high will
it rise? (c) How long after it leaves the hand will it return to the starting point? (d) When will its speed
be 16 m/s? Ans. (a) 3.06 s; (b) 46 m; (c) 6.1 s; (d) 1.43 s and 4.7 s
4,41
A bottle dropped from a balloon reaches the ground in 20 s. Determine the height of the balloon if (a) it
was at rest in the air and (b) it was ascending with a speed of 50m/s when the bottle was
dropped. Arts. (a) 1.96km; (b) 960m
/
at
the
md
the
the
Two balls are dropped to the ground from different heights. One is dropped 1/~s after the other, but
they both strike the ground at the same time, 5.0s after the first was dropped./(’a) What is the difference
in the heights from which they were dropped? (b) From what l)~ight was the first ball
dropped? Ans. (a) 62.5m; (b) 122.5 m
/!
A nut comes loose from a bolt on the bottom of an elevator as the elevator m moving up the shaft at
3 m/s. The nut strikes the bottom of the shaft in 2 s. (a) How far from the bottom of the shaft was the
42
UNIFORMLY ACCELERATED MOTION
[CHAP. 4
elevator when the nut fell off? (b) How far above the bottom was the nut 0.25 s after it fell
off? Arts. (a) 13.6m; (b) 14.0m
A marble, rolling with speed 20 cm/s, rolls off the edge of a table that is 80 cm high. (a) How long does
it take to drop to the floor? (b) How far, horizontally, from the table edge does the marble strike the
floor?. Arts. (a) 0.404s; (b) 8.tcm
A body projected upward from the level ground at an angle of 50" with the horizontal has an initial
speed of 40m!s. (a) How long will it take to hit the ground? (b) How far from the starting point wilt it
strike? (c) At what angle with the horizontal will it strike? Arts. (a) 6.3 s; (b) 161 m; (c) 50°
A body is projected downward at an angle of 30° with the horizontal from the top of a building 170m
high. Its initial speed is 40 m/s. (a) How long will it take before striking the ground? (b) How far from
the foot of the building will it strike? (c) At what angle with the horizontal will it strike?
Ans. (a) 4.2s; (b) 145m; (c) 60°
A hose lying on the ground shoots a stream of water upward at an angle of 40° to the horizontal. The
speed of the water is 20 m/s as it leaves the hose. How high up will it strike a wall which is 8 m
away? Ans. 5.4 m
A World Series batter hits a home run ball with a velocity of 40 m/s at an angle of 26° above the
horizontal. A fielder who can reach 3 m above the ground is backed up against the bleacher wall, which
is 110 m from home plate. The ball was 120 cm above the ground when hit. How high above the fielder’s
glove does the ball pass? Ans. 6.02 m
4,49 }/~Prov~ ~hat a gun will shoot three times as high when its angle or" elevation is 60° as when it is 30°, but the
,!f; t\b~allei/will carry the same horizontal distance.
~-’~ A bail is thrown upward at an angle of 30° to the horizontal and lands on the top edge of a building that
~is 20m away. The top edge is 5m above the throwing point. How fast was the ball
thrown? Arts. 20 th/s
/
A ball is thrown straight upward with a speed v from a point h meters above the ground. Show that the
time taken for the ball to strike the ground is (v/g)[1 + ~].
~
Chapter 2 Kinematics in
CHAPTER 2 REVI[EW QUEST]~ONS
For each of the multiple-choice questions below, choose the best answer.
Unless otherwise noted, use g = 10 m/s2 and neglect air resistance.
5. A bus starting from a speed of+24
m/s slows to 6 m/s in a time of 3 s. The
average acceleration of the bus is
(A) 2 m/s2
(B) 4 rrds2
(C) 6 m/s2
(D) - 2 n~Js2
(E) - 6 rds2
1. Which of the following statements is
tree?
(A) Displacement is a scalar and
distance is a vector.
(B) Displacement is a vector and
distance is a scalar.
(C) Both displacement and distance are
vectors.
(D) Neither displacement nor distance
are vectors.
(E) Displacement and distance are
always equal.
6. A train accelerates from rest with an
acceleration of 4 m!s2 for a time of 20 s.
What is the train’s speed at the end of 20
S?
(A) 0.25 m/s
(B) 4 m!s
(C) 2,5 m!s
(D) 0.8 m/s
(E) 80 m/s
2. Which of the following is the best
statement for a velocity?
(A) 60 miles per hour
(B) 30 meters per second
(C) 30 km at 45° north of east
(D) 40 km!hr.
(E) 50 km/hr southwest
7. A football player starts from rest 10
meters from the goal line and accelerates
away from the goal line at 5 m/s2. How
far from the goal line is the player after 4
3. A jogger runs 4 km in 0.4 hr, then 8
kava in 0.8 hr. What is the average speed
of the jogger?
(A) 10 krrdhr
(B) 3 kr~/hr
(C) 1 klrdhr
(D) 0.1 km/hr
(E) 100 km!hr
s?
(A) 6 m
(B) 30 m,
(C) 40 m
(D) 50 m
(E) 60 m
4. A motorcycle starts from rest and
accelerates to a speed of 20 m/s in a time
of 8 s. What is the motorcycle’s average
acceleration?
(A) 160 m!s2
(B) 80 m/s2
(C) 8 m!s2
(D) 2.5 m/s2
(E) 0.4 m/s2
28
Chapter 2 Kinematics in One Dimension~
8. A ball is dropped from rest. What is
the acceleration of the ball immediately
after it is dropped?
(A) zero
(B) 5 m!s2
(C) 10 m/s2
(D) 20 m/s2
(E) 30 n~/s2
12. Which two of the following pairs of
graphs are equivalent?
Questions 9-11:
A ball is thrown straight upward with a
speed of+12 m!s.
(h) x ~t
v0 ~t
9. What is the ball’s acceleration just
after it is thrown?
(A) zero
(B) 10 m/s2 upward
(C) 10 ~rds2 downward
(D) 12 m/s2 upward
(E) 12 m/s~ downward
(B)
V
(c)
10. How much time does it take for the
ball to rise to its maximum height?
(A) 24 s
(B) 12 s
(C) 10 s
(D) 2 s
(E) 1.2 s
(D)
11. What is the approximate maximum
height the ball reaches?
(A) 24 m
(B) 17 m
(E)
(c) m
(D) 7 m
5m
29
X
Chapter 2 Kinematics in One Dimension
Questions 13 - 14:
Consider the velocity vs. time graph
below:
13. A which time(s) is the object at rest?
(A) zero
(B) ~ s
(C) 3 s to 4 s
(D) 4 s only
(~) g s
14. During which interval is the speed of
the object decreasing?
(A)0to 1 s
(B) 1 s to 3 s
(C) 3 s to 4 s
(D) 4 s to 8 s
(E) the speed of the object is never
decreasing in this graph
3O
Chapter 2 Kinematics in One Dimensiot*
ANSWERS AND EXPLANATIONS TO CHAPTER 2 REVIEW QUESTIONS
Multlple Choice
1. B
Displacement is the straight-line length from an origin to a final position and includes
direction, whereas distance is simply length moved.
2. E
Velocity is a vector and therefore direction should be included.
3. A
Average speed is total distance divided by total time. The total distance covered by the
jogger is 12 km and the total time is 1.2 hours, so the average speed is 10 ki!!hr.
4. D
Av 20m/s
m
a=--2.5 ~
8s
s
At
5. E
vI-% 6m/s-24m/s
t
3s
m
6. E
v,f =v, +at : O+(4m/sZ)(2Os)= 80m/s
33
Chapter 2 Kinematics in One Dimension
7. D
X.f =Xo
1 2=(10m)+0+~(5~21(4s)2 =50m
2
8. C
The acceleration due to gravity is 10 m/s2 at all points during the ball’s fall.
~9. C
After the ball is thrown, the only acceleration it has is the acceleration due to gravity, 10
m!sa.
10. E
At the ball’s maximum height, vf= O. Thus,
vs =vo -gt=O
12rn/s
t=--=l.2s ~
lOm/s
ll.D
y=Tgt = 10 (1.2s)2=7.2m~7m
12. B
Both of thes~ graphs represent motion that begins at a high positive velocity, and slows
down to zero velocity.
13. B
The line crosses the axis (v = 0) at a time of 1 second.
14. A
The object begins with a high negative (backward) velocity at t = 0, then its speed
decreases to zero by a time of 1 s.
34
Chapter 3 Kinematics in Two Dimensions
CHAPTER 3 REVIEW QUESTIONS
For each of the multiple-choice questions below, choose the best answer.
Unless otherwise noted, use g = ] 0 m/s2 and neglect air resistance.
1. Whicl’~ of the following is NOT true of
a projectile launched from the ground at
an angle?
(A) The horizontal velocity is constant.
(B) The vertical acceleration is upward
during the first half of the flight, and
downward during the second half of
the flight.
(C) The horizontal acceleration is zero.
(D) The vertical acceleration is 10 m/s2,
(E) The time of flight can be found by
horizontal distance divided by
horizontal velocity.
2. A projectile is launched horizontally
from the edge of a cliff 20 m high with
an initial speed of 10 m/s. What is the
horizontal distance the projectile travels
before striking the level ground below
the cliff?
(A) 5 m
(B) I0 m
(C) 20 m
(D) 40 m
(E) 60 m
3. A projectile is launched from level
ground with a velocity of 40 m/s at an
angle of 30° from the ground. What wilt
be the vertical component of the
projectile’s velocity; just before it strikes
the ground? (sin 30° = 0.5, cos 30 =
0.87)
(A) 10 m/s
(B) 20 rrds
(C) 30 m/s
(D) 35 m/s
(E) 40 m!s
43
Chapter 3 Kinematics in Two Dime~sions
5. The acceleration in the x-direction
and the y-direction, respectively, are
(A) zero, 3 m/s2
(B) zero, 6 m/s2
(C) 5 rrds2, 3 m/s2
(D) 5 m!s2, 6 m/s2
(E) 5 ~rds2, 12 m/s2
Questions 4 - 6
A toy rocket moves in the horizontal
direction according to the equation x =
5t, and in the vertical direction according
to the equation y = 3t~, where x andy are
in meters and t is in seconds.
4. The length of the displacement vector
of the rocket from the origin (t = 0) at a
time of 2 s is most nearly
(A) 22 m
(B) 2 m
(C) - 2 m
(D) 250 m
(E) 16 m
6. The horizontal velocity after 10
seconds is most nearly
(A) zero
(B) 5 m/s
(C) 10 m!s
(D) 50 m/s
(E) 300 irds
\~’kKree Response Problem
~ Show all work in working the following question. The question i~r, th 10
points, a~the suggested time for answering the question is about 10 mi~s. The parts
within a qu~may not have equal weight.
~
1. (10 points) ~
~
Two planetary exploi~ land on an uncharted planet a~ecide t.o et st t_he_r_angy of
can~ boroug!,at a, lo .hg~ W~hen they fire a c~nno~l ’w~th a s~eed of }00 rr~/s at an
~7 fro _m the ho.rizN~al ground, t~e cannonball follows a parabolic
(~his unch~ planet.
(b) D ’~~ cannonba_ll reaches.
(~must have landed over a mile
~r (1 mile = 1600 m). _ o
(d) Tt~ at 100 m/s at an .a_ n.gl~ .of75 to
~ range x’ which is less than,
~efro a25 a~nchangle?
44
Chapter 3 Kinematics in Two Dimensions
ANSWERS AND EXPLANATIONS TO CHAPTER 3 REVIEW QUESTIONS
1. B
Since the vertical acceleration is due to gravity, it is always downward.
2. C
First we find the time of flight, which can be calculated from the height:
y=~gt2, sot=, 2~ ~=2s
Then, x = v~t = (lOm/s)(2s): 20m
3. B
Neglecting air resistance, the y-component of the velocity of the projectile just before it
lands is equal to the y-component of the velocity when it is first fired:
vy =(40m/s)sin30° ~ 20m/s
4. E
At a time oft = 2 s, x = 5(2s) = 10m and y = 3(2s)2 = 12m. Then the length of the
displacement vector can be found by the Pythagorean theorem:
irI:~+y~ :~f{10)~+(12)~ = 2~=16m
5. B
Both the horizontal and Vertical components of the displacement of the rocket at any time
1~
can be found by the general equation s = so + vot +~ at . Since the equation for x has no
t~ tenza, the horizontal acceleration must be zero. The vertical acceleration can be found
~
1~
.
by y=-at ,andsmcey=3 t,a=6~rds2.
2
6.8
"
Since x = vxt = 5t, then vx = 5 m/s, which remains constant.
45
12
~
~DER THE ACTION OF CONCURRENT FORCES
[CHAP. 2
THE COEFFICIENT OF KINETIC FRICTION (#k) is defined for the case in which
one surface is
sliding across another at constant speed. It is
friction force f
normal force Fn
THE COE~’~IICIENT OF STATIC FRICTION (#~) is defined for the case in which one surface is
just on the verge of sliding across another surface. It is
critical friction force f
normal force
FN
where the critical friction force is the friction force when the object is just on the verge of slipping.
Solved Problems
2.1
The object in Fig. 2-1(a) weighs 50 N and is supported by a cord. Find the tension in the
cord.
We mentally isolate the object for discussion. Two forces act on it, the upward pull of the cord and
the downward pull of gravity. We represent the pull of the cord by T, the tension in the cord. The pull
of gravity, the weight of the object, is w = 50 N. These two forces are shown in the free-body diagram of
Fig. 2-1(b).
(a)
(~,)
Fig. 2-1
The forces are already in component form and so we can write the first condition for equilibrium at
~ F~ = 0
~ F~ = 0
becomes
becomes
0=0
T- 50 N = 0
from which T = 50 N. Thus, when a single vertical cord supports a body at equilibrium, the tension in
the cord equals the weight of the body.
In Fig. 2-2(a), the tension in the horizontal cord is 30 N as shown. Find the weight of the
object.
As we saw in Problem 2.1, the tension in cord 1 is equal to the weight of the object hanging from it.
Therefore Ta = w, and we wish to find T~ or w.
Notice that the unknown force T~ and the known force of 30 N both pull on the knot at point P. It
therefore makes sense to isolate.the knot at P as our object. The free-body diagram showing the forces
on the knot is drawn as Fig. 2-2(b). The force components are also shown there.
We next write the first condition for equilibrium for the knot. From the free-body diagram,
~, Fx = 0
becomes
~,, F~ = 0
becomes
30 N - T,, cos 40° = 0
Tz sin 40° - w = 0
CHAP. 21
EQUILIBRIUMUNDER THE ACTION OF CONCURRENT FORCES
cord
13
T2 sih 40’
T2 cos 40~
cord
30 N
weight = w
(a)
(b)
Fig. 2-2
Solving the first equation for T~ gives T~ = 39.2 N. Substituting this value in the second equation gives
w = 25.2 N as the weight of the object.
A rope extends between two poles. A 90N boy hangs from it as shown in Fig. 2-3(a). Find
the tensions in the two parts of the rope.
We label the two tensions T~ and T2, and isolate the rope at the boy’s hands as the object. The
free-body diagram for the object is shown i~a Fig. 2-3(b).
10°
TI sin 10°~
T2¢os 5°
TI cos 10°
Tzsin 5°
x
w= 90N
Fig. 2-3
(b).
After resolving the forces into their components as shown, we can write the first condition for
equilibrium:
becomes
T~ cos 5° - T~ cos 10° = 0
~ F~ = 0
~ Fy = 0
becomes
T2 sin 5° + T~ sin 10° - 90 N = 0
When we evaluate the sines and cosines, these equations become
0.996T~ - 0.985T~ = 0 and 0.087T2 + 0.174T~ - 90 = 0
Solving the first for T2 gi~es ~*= 0.990T~. Substituting this in the second equation gives
0.086T~ + 0.174T~ - 90 = 0
from which T~ = 346 N. Then, because T2 = 0.990T~, we have T~ = 343 N.
A 200 N wagon is to be pulled up a 30° incline at constant speed. How large a force parallel to
the incline is needed if friction effects are negligible?
The situatio~ is shown in Fig. 2-4(a). Because the wagon moves at a constant speed along a straight
line, its velocity vectOr is constant. Therefore the wagon is in translational equilibrium, and the first
condition for equilibrium applies to it.
14
EQUILIBRIUM UNDER THE ACTION OF CONCURRENT FORCES
Support
[CHAP. 2
~orce
,~ Pulling
\
\
Weight
(a)
l?ig. 2-4
We igolate the wagon as the object. Three nonnegligible forces act on it: (1) the pull of gravity w
(its weight), directed straight down; (2) the force P exerted on the wagon parallel to the incline to pull it
up the incline; (3) the push Ev of the incline that supports the wagon. These three forces are shown in
the free-body diagram of Fig. 2-4(b).
For situations involving inclines, it is convenient to take the x-axis parallel to the incline and the
y-axis perpendicular to it. After taking components along these axes, we can write the first condition for
equilibrium:
becomes
P - 0.50w = 0
~ F~ = 0
becomes
F~v - 0.87w = 0
~, Fr = 0
Solving the first equation and recalling that w = 200 N, we find that P = 0.50w = 100 N. The required
pulling force is 100 N.
A 50 N box is slid straight across the floor at constant speed by a force of 25 N, as shown in
Fig. 2-5(a). (a) How large a friction force impedes the motion of the box? (b) How large is
the normal force? (c) Find ~k between the box and the floor.
Notice the forces acting on the box, as shown in Fig. 2-5(a). The friction force is f and the normal
force, the supporting force exerted by the floor, is FN. The free-body diagram and components are
shown in Fig. 2-5(b). Because the box is moving with constant velocity, it is in equilibrium. The first
condition for equilibrium tells us that
~F~=0 or 25 cos40°-f = 0
(a) We can solve for the friction force f at once to find that f = 19.2 N.
25 N
f
40*
[25 sin 40°
~
25 cos 40°
50 N
(b)
(a)
Fig. 2-5 -
CHAP. 2]
EQUILIBRIUM UNDER THE ACTION OF CONCURRENT FORCES
15
(b) To find FN we use the fact that
E~=o
FN + 25 sin 40° -- 50 = 0
or
Solving gives the normal force as Ft¢= 33.9 N.
(c) From the definition of it,, we have
f t9.2
~
’Findthe
tensions in the ropes shown in Fig. 2-6(a) if the supported object weighs 600 N.
Let us select as our object the knot at A because we know one force acting on it. The weight pulls
down on it with a force of 600 N, and so the free-body diagram for the knot is as shown in Fig. 2-6(b).
Applying the first condition for equilibrium to that diagram, we have
~F.=0
~ Fr = 0
or
T~cos 60°- T~ cos 60° = 0
T~ sin 60° + T~ sin 60° - 600 = 0
or
The first equation yields T~ = T~. (We could have inferred this from the symmetry of the system. Also
from symmetry, T~= T~.) Substitution of Ta for T~ in the second equation gives T~ =346N, and so
T2 = 346 N also.
Let us now isolate knot B as our object. Its free-body diagram is shown in Fig. 2-6(c). We have
already found that T~ = 346 N and so the equilibrium equations are
~ F. = 0
~ Fr = 0
or
or
T~ cos 20° - T~ - 346 sin 30° = 0
T~ sin 20° - 346 cos 30° = 0
The last equation yields T~ = 877 N. Substituting this in the prior equation gives Ts = 651 N. As stated
previously from symmetry T~ = ~, = 877 N. How could you have found T~ without recourse to symmetry?
[Hint: See Fig. 2-6(d).]
T5
(a)
x
600 N
T2
30°
30° /
(d)
Fig. 2-6
16
[CHAP, 2
EQUILIBRIUM UNDER THE ACTION OF CONCURRENT FORCES
n Ft’g.
2 7- 1s" in equilibrium. Find the normal force F~ in each case.
Each of the objectsi
w = 200N
w = 500 N
(c)
(a)
~ig. 2-7
We apply ~] Fr = 0 in each case.
(a)
FN + 200 sin 30° - 500 = 0
(b)
FN - 200 sin 30° -150 = 0
F~, - 200 cos 0 = 0
(c)
from which
from which
from which
FN = 400N
FN = 250 N
FN = (200 cos 0) N
For the situations of Problem 2.7, find the coefficient of kinetic friction if the object is movi
with constant speed, i.e., each object is in translational equilibrium.
We have already found FN for each case in Problem 2.7. To find f, the sliding-friction force, we use
~] F~ = 0. Then we use the definition of
(a) We have 200 cos 30° -f = 0 so that f = 173 N. Then,/z, =f/FN = 173/400 = 0.43.
(b) We have 200 cos 30° -f = 0 so that f = 173 N. Then,/~k =f/F,v = 173/250 = 0.69.
(c) We have -200 sin 0 +f = 0 so that f = (200 sin 0) N. Then, /~k =f/FN = (200 sin 0)/(200 cos 0) =
tan 0.
Suppose that in Fig. 2-7(c) the block is at rest. The angle of the incline is slowly increased. A
an angle 0 = 42°, the block begins to slide. What is the coefficient of static friction betwee
the block and the incline? (The block and surface are not the same as in Problems 2.7 and
2.8.)
At the instant the block begins to slide, the friction has its critical value. Therefore, N~ =f/FN at
that instant. Following the method of Problems 2.7 and 2.8, we have
F~ = w cos 0 and f = w sin 0
Therefore, when sliding just starts,
_ f _ wsin0_
gs-g-~-tan 0
But 0 was.found by e×periment to be 42°. Therefore, g~ = tan 42°= 0.90.
Pulled by the 8N block shown in Fig. 2-8(a), the 20N block slides to the right at constant
velocity. Find/~g between the block and the. table. Assume the pulley to be frictionless.
Because it is moving at constant velocity, the 20 N block is at equilibrium. Since the pulley is
frictionless, the tension in the continuous rope is the same on both sides of the pulley. Thus, we have
T~= T~= 8.00N.
Looking at the free-body diagram in Fig. 2-8(b) and recalling that the block is at equilibrium, we
CHAP. 2]
EQUILIBRIUM UNDER THE ACTION OF CONCURRENT FORCES
ta)
17
8N
have
or
or
f = ~=8.00N
FN = 20.0 N
Then, from the definition of ,u~,
f
8.00
20.0
F~
Supplementary Problems
2.11 i~} For the situation shown in Fig. 2-9, find the values of T~ and T~’if the object’s weight is
I~~ ~ 600 N. Ans. 503 N, 783 N
~600N
Fig. 2-9
The following coplanar forces p~fi on a ring: 200 N at 30°, 500 N at 80°, 300 N at 240°, and an unknown
force. Find the magnitude and direction of the unknown force if the ring is to be in
equilibrium. Ans. 350 N at 252°
In Fig. 2-10, the pulleys are frictionless and the system hangs at equilibrium. If wa, the weight of the
object on the right, is 200 N, what are the values of w~ and wz? Ans. 260 N, 150 N
t
Fig. 2-10
e
e
2.14
Fig. 2-11
Suppose w~ in Fig. 2-10 is 500N. Find the values of wx and ~ if the system is to hang in equilibrium as
shown. Ans. 288 N, 384 N
18
RENT FORCES
EQUILIBRIUM UNDER THE ACTION
[CHAP. 2
in Fig. 2-11 the friction between the block and the incline is negligible, how much must the object on
the right weigh if the 200 N block is to remain at rest? Arts. 115 N
in Fig. 2-11 remains at rest when w = 220 N. What are the magnitude and direction of the
friction force on the 200 N block? Arts. 105 N down the incline
the normal force acting on the block in each of the equilibrium situations shown in Fig.
Arts. (a) 34N; (b) 46N; (c) 91N
(c)
(b)
(a)
Fig. 2-12
The block shown in Fig. 2-12(a) slides with constant speed under the action of the force shown. (a) How
large is the retarding friction force? (b) What is the coefficient of kinetic friction between the block and
the floor? Ans. (a) 11.5N; (b) 0.34
The block shown in Fig. 2-12(b) slides at (onstant speed down the incline. (a) How large is the friction
force that opposes its motion? (b) What is the coefficient of sliding (kinetic) friction between the block
and the plane? Ans. (a) 38.6N; (b) 0.84
The block n Fig. 2-12(c) just begins to shde up the lnchne when the pushing force shown Is increased t.o
70N. (a) What is the critical static friction force on it? (b)What is the value of the coeffic,ent ot stat,c
friction? Ans. (a) 15N; (b) 0.165
2.21 If w = 40 N in the equilibrium situation shown in Fig. 2-13, find T~ and Tz.
Arts. 58 N, 31 N
T3
60°
~
Pig. 2-13
Fig. 2-14
2.22
Refer to the equilibrium situation shown in Fig. 2-13. The cords are strong enough to withstand a
maximum tension of 80N. What is the largest value of w that they can support as
shown? Ans. 55 N
2.23
The object in Fig. 2-14 is in equilibrium and has a weight w=80N. Find T~, Tz, Ts, and
T~. Ans. 37N, 88N, 77N, 139N
CHAP, 5]
NEWTON’S LAWS
45
Solved Problems
5.1 Find the weight on earth of a body whose mass is (a) 3 kg, (b) 200 g,
The general relation between mass m and weight w is w = mg. In this relation, m must be in
kilograms, g in m/s2, and w in newtons. On earth, g = 9.8 m/sz. The acceleration due to gravity varies
from place to place in the universe.
w = (3 kg)(9.8 m/s2) = 29.4 kg. m/s2 = 29.4 N
w = (0.20 kg)(9.8 m/s2) = 1.96 N
A 20 kg object that can move freely is subjected to a resultant force of 45 N in the
-x-direction. Find the acceleration of the object.
We make use of the second law in component form; )2 F~ = max, with )2 F~ = -45 N and m = 20 kg.
Then
)2 F~ ~45 N
2.25 N/kg = -2.25 m/s2
a, = .....
m 20 kg
where we have used the fact that 1 N = 1 kg. m/s~. Because the resultant force’ 6fi the obiect i~ in the
~-~,, -x-direction, its acceleration is also in that direction.
5.3 li~ A 5.0 kg object is to be given an upward acceleration of 0.30 m/s2 by a rope pulling straight
~’ upward on it. What must be the tension in the rope?
The free-body diagram for the object is shown in Fig. 5-1. The tension in the rope is T, and the
weight of the object is w =mg = (5.0 kg)(9.8 m/s2) = 49 N. Using )2 F~ = may with up taken as posiyive,
we have
T - mg= m% or T - 49 N = (5.0 kg)(0.30 m/sz)
from which T = 50,5 N. As a check, we notice that T is larger than w as it must be if the object is to
accelerate upward.
Fig. 5-1
Fig. 5-2
A horizontal force of 140 N is needed to pull a 60 kg box across the horizontal floor at
constant speed. What is the coefficient of friction between floor and box?
The free-body diagram for the box is shown in Fig. 5-2. Because the box does not move up or
down, ar = 0. Therefore,
~ T’y = may gives Fu -mg = (m)(0 m/sz)
from whicti we find that Fu = mg= (60kg)(9.8m/s~) =588N. Further, because the box is moving
horizontally at constant speed, ax = 0 and so
~, F, = tna~ gives
140 N -f = 0
NEWTON’S LAWS
46
[CHAP. 5
from which the friction force is f = 140 N. We then have
f 140 N
The only force acting on a 5 kg object has components E~ = 20 N and Fr = 30 N. Find the
acceleration of the object.
We make use of ~ Fx = max and ~ F~ = ma~, to obtain
a~= ZmE~=20N=4m/s
~
5 kg
EF~ 30N_6m~sz
These components of the acceleration are shown in Fig. 5-3. From the figure, we see that
a = ~-(4)" + (6)~ m/s~ = 7.21 m/sz
and 0 = arctan (6/4)=56°.
ax = 4 m/s~
Fig. 5-3
A 600 N object is to be given an acceleration of 0.70 m/s2. How large an unbalanced force
must act upon it?
Notice that the weight, not the mass, of the object is given. Assuming the weight was measured on
the earth, we use w =mg to find
w 600 N
m=~=~=61kg
Now that we know the mass of the object (61 kg) and the desired acceleration (0.70 m/sZ), we have
F = ma = (61 kg)(0.70 m/sz) = 43 N
A constant force acts on a 5 kg object and reduces its velocity from 7 m/s to 3 m/s in a time of
3 s. Find the force.
We must first find the acceleration of the object, which is constant because the force is constant,
Taking the direction of motion as positive, from Chapter 4 we have
a = v/- - vo = -4 m/s = _ 1.33 m/s~
3s
Now we can use F = m~t with m = 5 kg:
F = (5 kg)(-!.33 m/sz) = -6.7 N
The minus sign indicates that the force is a retarding force, directed opposite to the motion.
CHAP. 5]
NEWTON’S LAWS
47
A 400 g block with an initial velocity of 80 cm/s slides along a horizontal tabletop against a
friction force of 0.70 N. (a) How far will it slide before stopping? (b) What is the coefficient of
friction between the block and the tabletop?
(a) We take the direction of motion as positive. The only unbalanced force acting on the block is the
friction force, -0.70 N. Therefore,
~ F = ma becomes -0.70 N = (0.400 kg)(a)
from which a = -1.75 m/sz. (Notice that m is always in kilograms.) To find the distance the block
slides, we have vo = 0.80 m/s, v.~ = 0, and a = -1.75 m/s~. Then v~ - Voz = 2ax gives
@ - v~ (0 - 0.64) mZ/s
(b) Because the vertical forces on the block must cancel, the upward push of the table FN must equal
the weight mg of the block. Then
friction force
0.70 N
(0.40 kg)(9.8 m/s2)= 0.179
FN
A 600 kg car is moving on a level road at 30 m/s. (a) How large a retarding force (assumed
constant) is required to stop it in a distance of 70 m? (b) What is the minimum coefficient of
friction between tires and roadway if this is to be possible?
(a) We must first find the car’s acceleration from a motion equation. It is known that vo = 30 m/s,
v.e = 0, and x = 70 m. We use v} = vg + 2ax to find
900 m2/s2
a = -J = = -6.43 m/s2
2x
140 m
Now we can write
F = ma = (600 kg)(-6.43 m/s2) = -3860 N = -3.86 kN
(b) The force found in (a) is supplied as the friction force between the tires and rofidway. Therefore,
the magnitude of the friction force on the tires is f = 3860 N. The coefficient of friction is given by
~ =f/Fu, where F~ is th+ normal force. In the present case, the roadway pushes up on the car with
a force equal to the car’s weight. Therefore,
F~, = w =mg = (600 kg)(9.8 m/s~) = 5900 N
so that
#=~=~= .
f 3860 0 66
The coefficient of friction must be at least 0.66 if the car is to stop within 70 m.
An 8000 kg engine pulls a 40 000 kg train along a level track and gives it an acceleration
= 1.20 m/sz. What acceleration (a~) would the engine give to a 16 000 kg train?
For a given engine force, the acceleration is inversely proportional to the total mass. Thus
8000 kg + 40 000 kg 1 20 m sz 2 40 m s2
a~=--a!=~( .m~ / )= ’ /
As shown in Fig. 5-4(a), an object of mass m is supported by a cord. Find the tension in the
cord if the object is (a) at rest, (b) moving at constant velocity, (c) accelerating upward with
acceleration a = 3g/2 and (d) accelerating downward at a = 0.75g.
Two forces act on the object: the tension T upward and the downward pull of gravity rag. They are
shown in the free-body diagram of Fig. 5-4(b). We take up as the positive direction and write ~ F~ = ma~
in each case.
NEWTON’S LAWS
48
(a)
(b)
(c)
(d)
ay = 0:
ay = 0:
a~ = 3g/2:
ay =-~g/4:
T - mg = may = 0
T -mg = may = 0
T -mg = m(3g/2)
T - mg = m(-3g/4)
[CHAP. 5
or
or
or
or
T = mg
T= mg
T= 2.5rag
T= 0.25mg
Notice that the tension in the cord is less than mg in part (d); only then can the object have a downward
acceleration. Can you explain why T = 0 if ar = -g?
mg
~b)
]Fig. 5-4
A tow rope will break if the tension in it exceeds 1500 N. It is used to tow a 700 kg car along
level ground. What is the largest acceleration the rope can give to the car?
The forces acting on the car are shown in Fig. 5-5. Only the x-directed force is of importance,
because the y-directed forces balance each other.
~ F~ = max
becomes
1500 N = (’]00 kg)(a)
from which a = 2.14 m/s2.
fS.l’~ lCompute__~ the least acceleration with which a 45 kg woman can slide down a rope if the rope
~can withstand a tension of only 300 N.
The weight of the woman is mg = (45 kg)(9.8 m/sz) = 441 N. Because the rope can support only
300 N, the unbalanced downward force F on the woman must be at least 441 N- 300 N = 141 N. Her
minimum downward acceleration is then
a = _F = 141 N = 3.1 m/s2
m 45 kg
A 70 kg box is slid along the floor by a 400 N force as shown in Fig. 5-6. The coefficient of
friction between the box and the floor is 0.50 when the box is sliding. Find the acceleration of
the box.
Since the y-directed forces must balance,
FN =mg = (70 kg)(9.8 m/s:) = 686 N
49
NEWTON’S LAWS
But the friction force f is given by
f = #F,,, = (0.50)(686 N) = 343 N
Now write ~ F, = ma~, for the box, taking the direction of motion as positive:
400 N - 343 N = (70 kg)(a)
or a = 0.81 m/s2
as shown in Fig. 5-7, that a 70kg box is pulled by a 400N force at an angle of 30° to
horizontal. The coefficient of sliding friction is 0.50. Find the acceleration of the box.
400 N
f
346 N
Fig. 5-7
Because the box does not move up or down, we have ~ F~ = may = 0. From Fig. 5-7, we see that
this equation is
F,.v + 2OO N - mg = O
2)
But mg = (70 kg)(9.8 m/s = 686 N, and it follows that FN = 486 N.
We next find the friction force acting on the box:
f = !~FN = (0.50)(486 N) = 243 N
Now let us write ~ F~ = ma,~ for the box. It is
(346 - 243) N = (70 kg)(a,)
from which a~ = 1.47 m/s2.
A car moving at 20 m/s along a horizontal road has its brakes suddenly applied and
eventually comes to rest. What is the shortest distance in which it can be stopped if the
friction coefficient between tires and road is 0.90? Assume that all four wheels brake
identically.
The friction force at one wheel, call it wheel I, is
f~ = #Fro = !zwl
where w~ is the weight carried by wheel 1. We obtain the total friction force f by adding such terms for
all four wheels:
f = gw~ + #w2 + #w..~ + l.tw4 = lz(w~ + wz + w3 + w.O = l~W
where w is the total weight of the car. (Notice that we are assuming optimal braking at each wheel.) This
friction force is the only unbalanced force on the car (we neglect wind friction and such). Writing
F = ma for the car with F replaced by -ktw gives -~tw = ma, where m is the car’s mass and the positive
direction is taken as the direction of motion. However, w = mg; so the car’s acceleration is
a = -#w= - p.m~g = -/zg = (-0.90)(9.8 m/s") = -8.8 m/s~
m
m
We can find how far the car went before stopping by solving a motion problem, Knowing that
NEWTON’S LAWS
50
[CHAP. 5
v0 = 20 m/s, v,~ = 0, and a = -8.8 m/s2, we find from v~- vo~ = 2ax that
(0 - 400) m2/s2 22 7 m
- I7.6 m/sz
If the four wheels had not all been braking optimally, the stopping distance would have been longer.
~@~
As shown in Fig. 5-8, a force of 400N pushes on a 25kg box. Starting from rest, the box
achieves a velocity of 2.0 m/s in a time of 4 s. Find the coefficient of sliding friction between
box and floor.
We will need to find f by use of F = ma. But first we must find a from a motion problem. We know
that v0 = 0, vr = 2 m]s, t = 4 s. Using vr = Vo + at gives
a = v~. - Vo = 2 m/s = 0.50 m/sz
t 4s
Now we can write E ~ = max, where a, = a = 0.50 m]s~. From Fig. 5-8, this equation becomes
257 N -f = (25 kg)(0.50 m/sz) or f = 245 N
We now wish to use !z =f/FN. To find FN we write ~ F~ = mdy = 0, since no vertical motion occurs.
From Fig. 5-8,
FN -- 306 N - (25)(9.8) N = 0 or FN = 551 N
Then
f 245 ~ 44
~g. 5-8
Fig. 5-9
gol8 }) A 20 kg box s~ts on an mchne as shown ~n F~g. 5-9. The coefficient of shdmg friction between
~ box and incline is 0.30. Find the acceleration of the box down the incline.
In solving inclined-plane problems, we take x- and y-axes as shown in the figure, parallel and
perpendicular to the incline. We shall find the acceleration by writing P, F~ = ma,~ But first we must find
the friction force f
F~ = ma~, = 0 gives Fw - 0.87rag = 0
from which F~ = (0.87)(20 kg)(9.8 m/sz) = 171 N. Now we can find f from
f =/~F,~ = (0.30)(171 N) = 51 N
Writing ~ F~ = ma,, we have
f - 0.5rag = ma, or 51 N - (0.5)(20)(9.8) N = (20 kg)(a,)
from which a~ = -2.35 m/s~. The box accelerates down the incline at 2.35 m/sz.
>.5
CHAP. 5]
NEWTON’S LAWS
51
When a force of 500N pushes ona 25 kg box as shown in Fig. 5-10, the acceleration of the
box up the incline is 0.75 m/s2. Find the coefficient of sliding friction between box and incline,
box
een
The acting forces and their components are shown in Fig.’ 5-10. Notice how the x- and y-axes are
taken. Since the box moves up the incline, the friction force (which always acts to retard the motion) is
directed down the incline.
Let us first find f by writing ~ F~ = max. From Fig. 5-10, that gives
383 N -f - (0.64)(25)(9.8) N = (25 kg)(0.75 m/sz)
from which f = 207 N.
We also need FN. Writing ~ F~ = may = O, we have
FN -- 321 N - (0.77)(25)(9.8) N = 0 or
Then
# =~-~=5~07= 0.41
500 N
Fig. 5-10
Fig. 5-11
TWo blocks, of masses ml and m2, are pushed by a force P as shown in Fig. 5-11. The
coefficient of friction between each block and the table is 0.40. (a) What must be the value of
P if the blocks are to have an acceleration of 200cm/s2? (b) How large a force does rnI then
exert on m2? Use m~ = 300 g and m~ = 500 g,
The friction forces on the blocks are f~=0.4mag and fz=O.4m2g. We take the two blocks in
combination as the object for discussion; the horizontal forces on the object from outside (i.e, the
external forces on it) are P, f:, and f~. Although the two blocks do push on each other, the pushes are
internal forces; they are not part of the unbalanced external force on the two-mass object. For that
object,
~,, F~ = ma= becomes P -f~ -A = (ml + m~)a=
(a) Solving for P and substituting known values, we find
P = 0.4g(ma + mz) + (m~ + mz)a~ = 3.14 N + 1.60 N = 4.74 N
fnd
(b ) N~w c~nsider b~ck m~ a~ne. The f~rces a~ting ~n it in the x-directi~n are the push ~f b~ck m ~ ~n
it (which we represent by F~) and the retarding friction force f~ = 0.4m~g. Then, for it,
~ F= = ma=
z
We know that a, = 2.0 m/s and so
becomes
F~ - fz = mza~
F~ =fz + m2a= = 1.96N+ 1.00N = 2.96N
A cord passing over an easily turned pulley has a 7 kg mass hanging from one end and a 9 kg
mass hangmg from the other, as shown in Fig 5-12. (This arrangement is called A~wood s
machine." Find the acceleration of the masses and the tension in the cord.
52
NEWTON’S LAWS
[CHAP. 5
Because the pulley is easily turned, the tension in the cord wiil be the same on each side. The forces
acting on each of the two masses are drawn in Fig. 5-12. Recall that the weight of an object is rag.
It is convenient in situations involving objects connected by cords to take the direction of motion as
the positive direction. In the present case, we take up positive for the 7 kg mass, and down positive for
the 9 kg mass. (If we do this, the acceleration will be positive for each mass. Because the cord doesn’t
stretch, the accelerations are numerically equal.) Writing ~ F~ = mar for each mass in turn, we have
T-(7)(9.8)N=(7kg)(a) and (9)(9.8)N- T=(9kg)(a)
If we add these two equations, the unknown T drops out, giving
(9-7)(9.8)N=(16kg)(a)
z.
for which a = 1.23 m/s We can now substitute 1.23 m/s2 for a in either equation and obtain T = 77 N.
7(9.8) N
9(9.8) N
Fig. 5-12
Fig. 5-13
Fig. 5-13, the coefficient of sliding friction between block A and the table is 0.20. Also,
mA = 25’kg, m~ = 15 kg. How far will block B drop in the first 3 s after the system is released?
Since, for block A, there is no motion vertically, the normal force is
F~¢ = mAg = (25 kg)(9.8 m/s~) = 245 N
and
f= ~F~, = (0.20)(245 N) = 49 N
We must first find the acceleration of the system and then we can describe its motion. Let us apply
F = ma tO each block in turn. Taking the motion direction as positive, we have
T-f=mAa or T-49N= (25kg)(a)
and
mBg - T = m~a or -T + (15)(9.8) N = (t5 kg)(a)
We can eliminate T by adding the two equations. Then, solving for a, we find a = 2.45 m/s~.
Now we can work a motion problem with a = 2.45 m]s~, vo = 0, t = 3 s:
y = rot + ½at~ gives y = 0 + ½(2.45 m/s2)(3 s)2 = 11.0 m
as the distance B falls in the first 3 s.
How large a horizontal force in addition to T must pull on block A in Fig. 5-13 to give it an
acceleration of 0.75 m/s2 toward the left? Assume, as in Problem 5.22, that /z =0.20,
mA = 25 kg, and ra,~ = 15 kg.
If we were to redraw Fig. 5-13 for this case, we would show a force P pulling toward the left on A.
In addition, the retarding friction force f should be reversed in direction in the figure. As in Problem
5.22, f = 49 N.
We write F = ma for each block in turn, taking’the direction of motion to be positive. We have
P - T - 49 N = (25 kg)(0.75 m/s~) and T - (15)(9. 8) N = (15 kg)(0.75 m/s~)
NEWTON’S LAWS
CHAP. 5]
53
~
We solve the last equation for T and substitute in the previous equation. We can then solve for the
single unknown P, and we find it to be 226 N. /
The coefficient of static friction between a box and the fiat bed of a truck is 0.60. What is the
maximum acceleration the truck can have along level ground if the box is not to slide?
The box experiences only one x-directed force, the friction force. When the box is on the verge of
slipping, f = lz, w, where w is the weight of the box.
As the truck accelerates, the friction force must cause the box to have the same acceleration as the
truck; other~vise, the box will slip. When the box is not slipping, ~ ~ = max applied to the box gives
f = ma,. However, if the box is on the verge of slipping, f =/z,w so that #,w = max. Because w = rag,
this gives
ax =/t~rng = ~t,g = (0.60)(9.8 m/sz) = 5.9 m/sz
m
as the maximum acceleration without slipping.
¯
In Fig. 5-14, the two boxes have identical masses o~ 40 kg. Both experience a sliding friction
force with !z = 0.15. Find the acceleration of the boxes and the tension in the tie cord.
,rag
Fig, 5-14
Using f = ~FN, we find that the friction forces on the two boxes are
f~ = (0.15)(rag)
and f~ = (0. I5)(0.87rag)
But m =40t~g, sof,~ =59N andf~ =51N.
Let us now apply ~ F~ = max to each blodk in turn, taking the direction of motion as positive. This
gives
T - 59 N = (40 kg)(a)
and 0.5rag - T - 51 N = (40 kg)(a)
Solving these two equations for a and T gives a = L 08 m/s2 and T = 102 N.
5.26 In the system shown in Fig. 5-15(a), force F accelerates block ml to the right~ Find its
acceleration in terms of F and the coefficient of friction/~ at the contact surfaces.
. The horizontal forces on the blocks are shown in Fig. 5-15(b) and (c). Block mz is pressed against
m~ by its weight, m~g. This is the normal force where m~ and m~ are in contact, so the friction force
there is f = #m~g. At the bottom surface of rnz, however, the normal force is (ma + mz)g. Hence,
f’ = tz(mt + m~)g. We now write ~ F, = ma~ for each block, taking the direction of motion as positive:
T - #m~g = mza
and
F - T - Izmzg - !~(m~ + mz)g = m~a
NEWTON’S LAWS~
54
[CHAP. 5
(c)
Fig. 5-15
We can eliminate T by adding the two equations to obtain
F - 21tm2g - ~t(mt + mz)(g) = (m~ + mz)(a)
from which
5.27
F - 21zm~g
a=
In the system of Fig. 5-16, friction and the mass of the pulley are both negligible. Find the
acceleration of mz if m~ = 300 g, mz = 500 g, and F = 1.50 N.
Notice that ma has twice as large an acceleration as mz has. (When the pulley moves a distance d,
m~ move~ a distance 2d.) Also notice that the tension T~ in the cord pulling m~ is half T2, that in the cord
pulling the pulley, because the total force on the pulley must be zero. (F = ma tells us that this is so
because the mass of the pulley is zero.) Writing ~ F,. = ma~ for each mass, we have
T~ = (ml)(2a)
and F - Tz = mza
However, we know that Tt = ½T~ and so the first equation gives T~ = 4m~a. Substitution in the second
equation yields
F m~ 1.20 kg
1.50+ N0.50 kg = 0.882 m/s~
F = (4m~. + mz)(a) or a = 4m~ +
F
Fig. 5-16
5.28
30O N
Fig. 5-17
In Fig. 5-17, the weights of the objects are 200N and 300N. The pulleys are essentially
frictionless and massless. Pulley Pt has a stationary axle, but pulley Pz is free to move up and
down. Find the tensions T~ and Tz and the acceleration of each body.
Mass B wil! rise and mass A will fall. You can see this by noticing that the forces acting on pulley Pz
are 2Tz up and T~ down. Since the pulley has no mass, it can have no acceleration, and so T~ = 2Tz (the
inertialess object transmits the tension). Twice as large a force is pulling upward on B as on A.
CHAP. 5]
NEWTON’S LAWS
@~
55
Let a be the downward acceleration of A. Then a/2 is the upward acceleration of B. (Why?) We
now write ~ Fy = mar for each mass in turn, taking the direction of motion as positive in each case. We
have
T~ - 300 N = (m~)(½a) and 200 N - Ta = mAa
But m = w/g and so mA = (200/9,8) kg and mB= (300/9.8) kg. Further, T~ = 2Tz, Substitution of these
values in the two equations allows us to compute T~ and then T~ and a. The results are
T~ = 327 N T2 = 164 N a = 1.78 m/s~
weight of the object is equal to rag. It is also equal to the gravitational force G(Mm)/rz, where r is the
earth’s radius, Hence,
Mm
from which
~ gr~ (9.8m/s~)(6.37x106m)2 .... 2~.
Supplementary Problems
5.30
A force acts on a 2 kg mass and gives it an acceleration of 3 m/sZ:~Nhat acceleration is produced by the
same force when acting on a mass of (a) 1 kg? (b) 4kg? (c) How large is the force?
Arts. (a) 6m/s-’; (b) L5m/sZ; (c) 6N
5.31
An object has a mass of 300 g. (a) What is its weight on earth? (b) What is its mass on the moon? (c)
What will be its acceleration on the moon when a 0.50 N resultant force acts on it?
Arts. (a) 2.94N; (b) 0.300kg; (c) 1.67m/sz
gone? Ans. (a) 3.2s; (b) i2.8m
~
p
of 8 m/s? (b) How far will it have
A 900 kg car is going 20 m/s along a level road. How large a constant retarding force is required to stop
it in a distance of 30 m? (Hint: First find its deceleyation.) Arts. 6000 N
A 12.0g bullet is accelerated from rest to a speed of 700m/s as it travels 20cm in a gun barrel.
Assuming the acceleration to be constant, how large was the accelerating force? (Be careful of
uniis). Ans. 14.7 kN
A 20 kg crate hangs at the end of a long rope. Find its acceleration when the tension in the rope is (a)
250N, (b) 150N, (c) zero, (d) 196N.
Ans. (a) 2.7 m/s" up; (b) 2.3 m/s~ down; (c) 9.8 m/sz down; (d) zero
A 5 kg mass hangs at the end of a cord. Find the tension in the cord if the acceleration of the mass is (a)
1.5 m/s~ up, (b) 1.5 m/s~ down, (c) 9.8 m/s~ down. Arts. (a) 56.5 N; (b) 4!.5 N; (c) zero
A 700 N man stands on a scale on the floor of an elevator. The scale records the force it exerts on
whatever is on it. What is the scale reading if the elevator has an acceleration of (a) 1.8 m/sz up? (b)
1.8 m/s~ down? (c) 9.8 m/s~ down? Arts. (a) 829 N; (b) 571 N; (c) zero
NEWTON’S LAWS
56 .
[CHAP. 5
Using the scale described in Problem 5.37, a 65 kg astronaut weighs himself on the moon, where
g = 1.60 m/s2. What does the scale read? Ans. 104 N
A cord passing over a frictionless, massless pulley has a 4k~, object tied to one end and a !2kg object
tied to the other. Compute the acceleration and the tension in the cord. Arts. 4.9 m/s2, 59 N
An elevator starts from rest with a constant upward acceleration. It moves 2.0m in the first 0.60s. A
passenger ix the elevator is holding a 3 kg package by a vertical string. What is the tension in the string
during the accelerating process? Ans. 63 N
Just as her parachute opens, a 60 kg parachutist is failing at a speed of 50 m/s. After 0.80 s has passed,
the chute is fully open and her speed has dropped to 12.0 m/s. Find the average retarding force exerted
upon the chutist during this time if the deceleration is uniform. Arts. 2850 + 588 = 3438 N
5.42
A 300 g mass hangs at the end of a string, A second string hangs from the bottom of that mass and
supports a 900 g mass, (a) Find the tension in each string when the masses are accelerating upward at
0.70 m/sz. (b) Find the tension in each string when the acceleration is 0.70 m/s~ downward.
Ans. (a) 12,6Nand9.45N;(b) 10.9 N and 8.19 N
A 20 kg wagon is pulled along the level ground by a rope inclined at 30° above the horizontal. A friction
force of 30N opposes the motion. How large is the pulling force if the wagon is moving with (a)
constant speed and (b) an acceleration of 0.40 m/sZ? Ans. (a) 34.6 N; (b) 43.9 N
A 12kg box is released from the top of an incline that is 5.0m long and makes an angle of 40° to the
horizontal~ A 60 N friction force impedes the motion of the box. (a) What will be the acceleration of the
box and (~) hqw long will it take to reach the bottom of the incline? Ans. (a) 1.30 m/sZ; (b) 2.8 s
@~.
For t~e situation outlined in Problem 5.44, what is the coefficient of friction between box and
~o
incline? Am. 0.67
An inclined pane makes an angle of 30 w th the horizontal. Find the constant force, applie&parallel to
the plane, required to CaUse a 15 kg box to slide (a) up the plane with acceleration 1.2 m/s2 and (b)
down the incline with acceleration 1’.2 m/sz. Neglect friction forces. Ans, (a) 91.5 N; (b) 55.5 N
A horizontal force P is exerted on a 20 kg box to slide it up ~ 30° incline. The friction force retarding the
motion is 80 N. How large must P be if the acceleration of the moving box is to be (a) zero and (b)
0.75 m/sZ? Ans. (a) 206 N; (b) 223 N
An inclined plane making an angle of 25° with the horizon(a! has a pulley at its top, A 30 kg block on the
plane is connected to a freely hanging 20 kg block by means of a cord passing over the pulley, Compute
the distance the 20 kg block will fall in2 s starting from rest. Neglect friction. Ans. 2.87 m
Repeat Problem 5.48 if the coefficient of friction between block and plane is 0.20.
Ans. 0.74 m
A horizontal force of 200N is required to cause a 15kg block to slide up a 20° incline with an
acceleration of 25 cm/sz. Find (a) the friction force on the block and (b) the coefficient of
friction. Ans. (a) 134N; (b) 0.65
Find the acceleration Of the blocks in Fig. 5-18 if friction forces are negligible. What is the tension in the
cord connecting them? Arts. 3.33 m/sz, 13,3 N
4.0 kg
5.0 kg F=30 N
Fig. 5-18
CHAP. 5]
NEWTON’S LAWS
57
Repeat Problem 5.51 if the coefficient of friction between the blocks and the table is
0.30.
Arts. 0.393 m/s2, 13.3 N
ct
5°53
How large a force F is needed in Fig. 5-19 to pull out the 6.0 kg block with an acceleration of 1.50 m/s2 if
the coefficient of friction at its surfaces is 0.40? Arts. 48.2 N
F
nd
Fig. 5-19
on
he
i/~In Fig. 5-20, how large a force F is needed to give the blocks an acceleration of 3.0 m/s2 if the coefficient
of friction between blocks and table is 0.20? How large a force does the 1.50 kg block then exert on the
2.0 kg block? Arts. 22.3 N, 14.9 N
~
~nd
tO
(b)
the
(b)
t:igo 5-20
5.56
(a) What is :the smallest force parallel to a 37° incline needed to keep a 100 N weight from sliding down
the incline if the coefficients of static and kinetic friction are both 0.30? (b) What parallel force is
required to keep the weight moving up the incline at constant speed? (c) If the parallel pushing force is
94 N, what will be the acceleration of the object? (d) If the object in (c) starts from rest, how far will it
move in 10 s? Ans. (a) 36 N; (b) 84 N; (c) 0~98 m/sz up the plane; (d) 49 m
A 5 kg block rests on a 30° incline. The coefificient of static friction between the block and the incline is
0.20. How large a horizontal force mus,~push on the block if the block is to be on the verge of sl)ding (a)
up the incline and (b) down the incline. Ans. (a) 43N; (b) 16.6 N
~-~ Three blocks with masses 6 kg, 9 kg, and 10 kg are connected as shown in Fig. 5-21. The coefficient of
y friction between the table and the 10 kg block is 0.2. Find (a) the acceleration of the system and (b) the
~ tension in the cord on the left and in the cord on the right. Arts. (a) 0.39.m/s2; (b) 61 N, 85 N
the
ute
an
Fig. 5-21
the
radius is about 6370 km. An object that has a mass of 20 kg is taken to a height of 160 km
above the earth’s surface. (a) What is the object’s mass at this height? (b) How much does the object
weigh (i.e., how large a gravitational force does it experience) at this height?
Ans. (a)20kg; (b) 186.5N
The radius or: the earth is about 6370 kin, while that of Mars is about 3440 km. If an object weighs 200 N
on earth, what would it weigh, and what would be the acceleration due to gravity, on Mars? The mass of
Mars is 0.11 that of earth. Arts. 75 N, 3.7 m/st
Chapter 4 Forces and Newton ~ Laws of Motion kv {~
CHAPTE~ 4 ~EV~EW QUESTIONS
For each of the multiple-choice questions below, choose the best answer.
Unless otherwise noted, use g = 10 m/s2 and neglect air resistance.
1. The amoun{ of force needed to keep a
0.2 kg hockey puck moving at a constant
speed of 7 m!s on frictionless ice is
(A) zero
(B) 0.2 N
(O 0.7 N
(D) 7 N
Ta
(E) 70 N
=30N
4. Which of the following is true of the
magnitudes of tensions T1, T2, and T3 in
the ropes in the diagram shown above?
(A) T~ must be greater than 30 N.
(B) The tension T2 is greater than TI.
(C) The y-component of T2 and T3 is
equal to 30 N.
(D) The sum of the magnitudes of T2 and
T3 is equal to %.
(E) The sum of the magnitudes of T~ and
2. A force of 26 N is needed to
overcome a frictional force of 5 N to
accelerate a 3 kg mass across a floor.
What is the acceleration of the mass?
(A) 4 m/s2
(B) 5 m/s2
(C) 7 1TI/S2
(D) 20 m/s2
(E) 60 m/s2
T2 is equal to T3.
3. A force of 1.00 N directed at an angle
of 45° from the horizontal pulls a 70 kg
sled across a frozen frictionless pond.
The acceleration of the sled is most
nearly (sin 45° = cos 45° = 0.7)
(A) 1.0 m/s2
(B) 0.7 m/s2
(C) 7 m!s2
(D) 35 m/s2
(E) 50 m/s2
5. Two blocks of mass m and 5m are
connected by a light string which passes
over a pulley of negligible mass and
friction. What is the acceleration of the
masses in terms of the acceleration due
to gravity,, g?
(a)4g
(B) 5 g
(O 6 g
(D) 4/5 g
(E) 2/3 g
64
Chapter 4 Forces and Newton ’s Laws of Motion
9. Which of the following diagrams of
two planets would represent the largest
gravitational force between the masses?
6. A 1-kg block rests on a frictionless
table and is connected by a light string to
another block of mass 2 kg. The string is
passed over a pulley of negligible mass
and friction, with the 2 kg mass hanging
vertically. What is the acceleration of
the masses?
(A) 5 m/s2
(B) 6.7 m!s2
(C) 10 m!s2
(D) 20 ~rds2
(E) 30 m!s2
(A)
rn
(B) m
©
©
"~
7. Friction
(A) can only occur between two surfaces
which are moving relative to one
another.
(B) is equal to the normal force divided
by the coefficient of friction.
(C) opposes the relative motion between
the two surfaces in contact.
(D) only depends on one of the surfaces
in contact.
(E) is always equal to the applied force.
2r
~"
2m
(C) m
2r
~
~
2111
(D) m
©
~----- r ---~
(E)
2m
2m
30°
10. A satellite is in orbit around the
earth. Consider the following quantities:
I. distance from the center
of the earth
II.
mass of the earth
mass of the satellite
III.
8. A 2-kg wooden block rests on an
inclined plane as shown above. The
frictional force between the block and
the plane is most nearly (sin 30 = 0.5,
cos 30 = 0.87, tan 30 = 0.58)
(A) 2 N
The gravitationa! acceleration g depends
on which of the above?
(A) i only
(B) I and II only
(C) III only
(D) I and III only
(E) I, II, and III
03) l0
(C) 12 N
(D) 17 N
(E) 20 N
65
Chapter 4 Fo ces a ~d Newton’s Laws of Motion
(c) A~ he block begins to move, the same force determined in part (b)
on the !~,~. What is the accellration of the block?
act
/ 0 = 20° and h~oefficient of kinetic friction gk = 0. I.
(d) ~e/~ ’~.ow tripled to 3F, which then pulls the block u, an ~i~c of angle
J
i. Di~aw the free-bb~or the blilk as it is
,ulled up the incline.
acting on the block as it
ii. Determine the
slides up the
e acceleration of the block as it is
~ the incline.
ANSWERS AND EXPLANATIONS TO CHAiI~TER 4 REVIEW QUESTIONS
1.A
The law of inertia states that no net force is needed to keep an object moving at a constant
velocity.
2. C
The net force is 26 N - 5 N = 21 N, and the acceleration is
a = F,,~, _ 21N = 7m/s~
m 3kg
3. A
Only the x-component of the force accelerates the sled horizontally: F~~ (I00 N)cos45° =
F~ 70 N
Im/s~70 N. Then a ....
m 70kg
67
Chapter 4 Forces and Newton’s Laws of Motion
,~k~Q ~ )
4. C
T~y + T3y must equal the weight of the block (30 N), since the system is m equilibrium.
5. E
The net force acting on the system is 5mg -mg = 4 rag. Then
F,et = 4 mg= (5 + 1)ma , and a = ~-g.
6. B
The weight of the 2-kg block is the net force accelerating the entire system.
2
2mg = 3ma, so a = -~g = 6.7 m/s2.
7. C
Friction acts on each of the surfaces in contact that are moving or have the potential for
moving relative to each other.
8. B
Since the block is in static equilibrium, the frictional force must be equal and opposite to
the component of the weight pointing down the incline:
f =mgsin0 = (2kg)00m/s2)sin30= 10N
9. D
The greater the mhss, and the smaller the separation distance r, the greater the force
according to Newton’s law of universal gravitation.
10. B
Since the acceleration due to gravity g -GM~
r2 , it does not depend on the mass of the
satellite.
~ Response Problem Solution
~
New__lope of the Fvs. a graph. So,
(b) 2 points /
~
T.he ~ceF’~comes the maximum static
~2rcflS:FN : /t ,mg : (O.2)(3 kg)(t Om/ s2 ): 6 N~
68
Chapter 5 Dynamics of Un~orm Circular Motion
~¯
CI~APTER 5 RE¥IEW QUESTIONS
For each of the multiple choice-questions below, choose the best answer.
Unless o~hervvise noted, use g = l O m/s2 and neglect air resistance.
1. A ball on the end of a string is swung
in a horizontal circle, rotating clockwise
as shown. When the ball is at a
particular poiiat in the circle, the
direction of the velocity, centripetal
force, and centripetal acceleration
vectors respectively are shown below:
Which of the fol!ov~ing best represents
the position of the ball as it rotates
clockwise?
Top View
(A)
,."---".,
(B)
2. If the string were suddenly cut when
the ball is at the position shown in
answer (E) in the previous question, the
subsequent motion of the ball would be
(A) to move to the right.
(B) to move to the left.
(C) to move to the top of the page.
(D) to move down and to the right
(E) to move up and to the left.
A 40 kg child sits on the edge of a
carnival ride at a radius of 4 m.
The ride makes 3 revolutions in 6 s.
3. The period of revolution for this ride
is
(A) ~/~ rev/s
(B) ~ s
(C) 2 rev/s
(D) 2 s
(E) 4 s
4. The speed of the child is most nearly
(A) 4 m/s
(B) 12 m/s
(C) 24 m!s
(D) 120 m/s
(E) 360 m/s
5. The force which is holding the child
on the ride is most nearly
(a) 30 N
(B) 160 N
(C) 320 N
(D) !440 N
(E) 2880 N
(D)
75
Chapter 5 Dynamics of Uniform Circular Motion
7. The car has a mass m and a speed v as
it moves around the track of radius R.
Which of the following expressions can
be used to find the value of the
coefficient of friction between the tires
and the road?
Questions 6- 7:
The figure below shows the top view of
a car going ’around a horizontal circular
track at a constant speed in the
counterclockwise direction. Assume the
frictional force between the tires and the
road is at its maximum value.
(B) # =
R
(c) # = ~gR
122
6. Which of the following vectors
represents the frictional force acting on
the tires of the car?
(D) # =-gR
(E) # =R
(A) ~
8. The acceleration due to gravity g at a
distance r from the center of a planet of
mass Mis 9 m!s2. In terms of the orbital
distance r, what would the speed of this
satellite have to be to remain in a
circular orbit arbund this planet at this
distance?
(B)
(C)
(B) v = 3r
(C) v = 9~rr
(D) v = 9r
(E) ~ = 33~r
76
Chapter 5 Dynamics of Uniform Circular Motion
ANSWERS AND EXPLANATIONS TO CHAPTER 4 REVIEW QUESTIONS
Choice
1. B
The velocity vector is always tangent to the path and in the direction of motion, and the
centripetal force and acceleration are toward the center of the circle.
2. D
At the instant the ball is in the ppsition shown in answer (E) in the previous question, it is
moving tangent to the path, down and to the right.
3. C
6s
Period T =-- = 2s .
3 rev
4. B
2,~ 2~’(4 m)
v=--=12m/s
T
2s
’,
/ v
5. D
F =mv2 = (40kg)(12m/s)2 = 1440N
r
4m
6. B
The frictional force provides the centripetal force that causes the car to move in a circle.
7. D
f = ,uFN -
R
, and Fu = rag, so/!mg -
R
, mad # = -gR
8. A
From section 5.6 in the textbook, v = ~ = ~rr = 3~r
pler’s 3ra law~-’ff6~ in the textbook) state~quare of the orbital~iod is ~
p~oportionalJ~he cube oftl~orbital radius~y a ~~ lo er orb"
79
86
ANGULAR MOTION IN A PLANE @
[CHAP. 9
RELATI
ANGULAR AND TANGENT~AL~~IITIIES~ When a wheel of
radius r rotates on its a’~a point on the rim of the.~gheel is described in terms of the
circumfere~ial distance s it hR~..oved, its tangential s~d v, and its tangential acceleration a.
These qua~d to the a’~l~r quantities 0~, and o:, which describe the rotation of the
wheel, through th re~
" "~ J
s=
a=
provided r~ple reasoning; s can be shown to be the
length of b~e whe~uld roll (without slipping) if free to
do so. In ~ and accelerat~f a point on the belt or of the
center of the wheel
/
"
CENTRIPETAL ACCELERATION: A point mass m moving with constant speed v around a
circle of radius r is undergoing acceleration. Although the magnitude of its linear velocity is not
changing, the direction of the velocity is continually changing. This change in velocity gives rise to an
acceleration ac of the mass, directed toward the center of the circle. We call this acceleration the
centripetal acceleration ; its value is given by
2 z
(tangential
speed)
a~ = radius
of circular
path = v~where v is the speed of the mass around the perimeter of the circle.
Because v = {or, we also have a¢ = (.o2r, where a~ must be in radian measure.
THE CENTRIPETAL FORCE is tbe unbalanced force that must act on the mass m moving in its
circular pa{h of radius r to give it the centripetal acceleration v~/r. From F = ma, we have
mn2
Required centripetal force = F~ = -- = moat
r
It must be directed toward the center of the circular path.
Solved Problems
9.1
Express each of the following in terms of the other angular measures: (a) 28°, (b) ~ rev/s, (c)
o
1 rev
28 = (28 deg)(36~eg) = 0.078 rev
lrev
2 rev 360deg
~V=(0" 5 ~-)(~) = 90~-
deg
CHAP, 91
89
ANGULAR MOTION IN A PLANE
given by
~o mnst be
on the ~
mass with~
by
cord
What is the maximum speed at which a car can round a curve of 25 m radius on a level road if
of static friction between the tires and road is 0.80?
The radial force required to keep the car in the curved path (the centripetal force) is supplied by the
force of friction between thb tires and the road. If the mass of the car is m, then the maximum friction
(and centripetal) force is 0.80rag; this arises when the car is on the verge of skidding sideways.
Therefore, the ma.ximmn speed is given by
-- = 0.80rag
r
or v = 0.~ = ~/(0.80)(9.8 m/sZ)(25 m) = 14 m/s
~gr
spaceship orbits the moon at a height of 20 000 m. Assuming it to be subject only to the
avitational pull of the moon, find its speed and the time it takes for one orbit. For the
moon, m,~ = 7.34 × 107z kg and r = 1.738 × 106 m.
The gravitational force of the moon on the ship supplies the required centripetal force:
where R is the radius of the orbit. Solving, we find that
~ = ,,/(6.67 × 10-~’ N. mZ/kg2)(7.34 × 10-~z1.67km/s
kg)=
v =~’ R~
(1.738 + 0.020) × 106 m
from which we find that
Time for one orbit = 2zR= 6.62 × 103s = 110 min
es shown in Fig. 9-3, a ball B is fastened to one end of a 24 cm string, and the other end is
ld fixed at point O. The ball whirls in the horizontal circle shown. Find the speed of the ball
in its circular path if the string makes an angle of 30° to the vertical.
Fig. 9-3
[CHAP. 9
ANGULAR MOTION IN A PLANE
90
The only forces acting on the bal! are the bali’s weight mg and the tension T in the cord. The
tension must do two things: (1) balance the weight of the ball by means of its vertical component,
T cos°30°; (2) supply the required centripetal force by means of its hm’izontal cm,nponent, T sin 30°.
Therefore we can write
T cos 30° =mg and T sin 30° = -Solving for T in the first equation and substituting it in the second gives
mg sin 30° = rnv__~~ or v = ~frg(O. 577)
cos 30° r
However, r = ~ = (0.24 m) sin 30° = 0.12 m and g = 9.8 m/s2, from which v = 0.82 m/s.
cm
As sh~n in Fig. 9-4, a 20 g bead slides from rest at A along a frictionless wire. 25and
and R is’~5~m, how large a force must the wire exert on the bead when it is at
(b) point
B
h
mg
Fig.
B. It has fallen through a distance h - 2R and so its
(a) Let us first find the speed of the bead at
loss in GPE is mg(h - 2R). This must (
hwhere v is the speed of the bead
B.
15 m) = 1.715 m/s
v = ~ - 2R)
As shown in Fig. 9-4(b), t~ forces act on the
~ is at B: (!) the weight of the bead mg
and (2) the (assumed do/nward) push P of the wire on ~e ~ad. Together, these two forces must
supply the required intripetat force, mvZ/R, if the bead ~to follow the circular path. We
therefore write
/mg + p = m--~-~
or
, = ~-~- mg= (0.02 kg)[(-~ ~--
/
ar path.
The w~e must exert a 0.98 N downward force on the bead to hold it in a c
Th~s centripetal
similar atforce.
pointTherefore
D, but now
the alone
weight
is furnish
perpendicular ;he direction of the
(b) iqmred
the wire
must
~s before, we have
v = X~g(h - R) = ~0.20 m) = 1.98 m/s
and
my
P=R
0.050 m
1.57 N
CHAP. 9]
~
ANGULAR MOTION IN A PLANE
9t
As shown in Fig. 9-5, a 0.9kg body attached to a cord is whirled in a vertical circle of radius
2.5 m. (a) What minimum speed v, must it have at the top of the circle so as not to depart
from the circular path?
o~r~ete’? (c) Find the tension Tb in the cord when the body is at the bottom of the
circle and mgving with the critical speed Vb.
As Fig. 9-5 shows, two radial forces act on the object at the top: (1) its weight tng and (2) the
tension T. The resultant of these two forces must supply the required centripetal force.
-- = mg + T,
r
For a given r, v will be smallest when T, = 0. In that case,
my,
Using r = 2.5 m and g = 9.8 m/s2 gives v, = 4.95 m/s as the speed at the top.
/~NIn traveling from bottom to top, the body rises a distance 2r. Therefore, with v, = 4.95 m/s as the
speed at the top and with vb as the speed at the bottom, conservation of energy gives
KE at bottom = KE at top + GPE at top
~.mvb = ½myz, + mg(2r)
where we have chosen the bottom of the circle as the zero level for GPE. Notice that m cancels.
Using v, = 4.95 m/s, r = 2.5 m, and g = 9.8 m/s2 gives vb = 11.1 m/s.
When the object is at the bottom of its path, we see from Fig. 9-5 that the unbalanced radial force
on it is T~ - rag. This force supplies the required centripetal force:
T~ -mg = -Using m = 0.9 kg, g = 9.8 m/s2, v~, = 11.1 m/s, and r = 2.5 m gives
7; =re(g+@) =53N
A curve of radius 30 m is to be banked so that a car may make the turn at a ~peed of 13 m/s
without depending on friction. What must be the slope of the curve (the banking angle)?
The situation is shown in Fig. %6 if friction is absent. Only two forces act upon the qar: (1) the
weight rag of the car and (2) the normal force F~ exerted by the pavement on the car.
The force FN must do two things: (1) its vertical component, FN COS O, must balance the car’s weight;
(2) its horizontal component, FN sin 0, must supply the required centripetal force. We can therefore
ANGULAR MOTION IN A PLANE
92
[CHAP, 9
sin 0
radius = 30 m
Fig. 9-6
write
FN cos 0 = mg
and
FN sin 0 = --
Dividing the second equation by the first causes FNand m to cancel and gives
o2 (13m/s)~
0.575
tan 0 ......
gr (9.8m/s~)(301n)
From this we find that O, the banking angle, must be 30°.
~
As’stlpwn in Fig. 9-7, a cylindrical shell of inner radius r rotates at angular speedfA
wooden~ock rests against the inner surface and rotates with it. If the coefficient o~icti~n
abetween b]t~ and surface is ~, how fast must the shell be rotating if the block i~ot to slip
p~~ck so it ~viil not
/ ~--~143 rev/s
N
-- v (0.30)(1.50 m)
ANGULAR MOTION IN A PLANE
CHAP, 9]
93
Supplementary Pr~Me~ns
9.17
Convert (a) 50 r~ to radians, (b) 48~g rad to revolutions, (c) 72 rps to rad/s, (d) !500 rp~to rad/s, (e)
22 rad/s to rpm, (]N)X2 rad/s to deg/s.
Arts. (a) 314 rad; (b~,24 rev; (c) 452 rad/s; (d) 157 rad/s; (e)210 rev/min; (/) ll~deg/s
9.18
Express the angular speed~4~ deg/s in (a) rev/s, (b) rev/min, and (c) rad/~
Ans. (a) O.IlI rev/s; (b) 6.~ev/min; (c) 0.698 rad/s
9.19
A
flywheel
480
rpm. A~s~Orad/s,
Cor~p, ute the
angular speed at any po~t on the wheel and the tangential
speed
30 cmturns
fromatthe
center.
15.1m/s~
9o2~
It is desired that the outer edge of a ~g~inding wheel 9.j~n ~in radius move at a rate of 6 m/s. (a)
Determine the angular speed of the wheel.~) What le~h o(thread ~could be wound on the rim of the
wheel in 3s when it is turning at tlg~~ (~)1?~_~ ~,
9.21
,
Through how many radians does ~rth’s surface move in6h as a result of the earth s
rotation? What is the sp.eed of a point/n th~equator? Take the radius of the earth to be
6370km. Ans. 1.57rad, 463m~/
~
,
9.22 A wheel 25 cm in radius turning at)40 rpm ~ncreases its s~e~ed to 660 rprrUm 9 ~ Find (a) the constant
angular acceleration in rev/s2 a);~ in rad/s2~cceleration of a point on its
rir~. Arts. (a) lrev/s’-=6,~, rad/sZ; (b) i57cm!s
9.Z3
The angular speed of a~hsk decreases uniformly from 12 to 4ra~s in 16s. Compute’the angular
acceleration and the~ber of revolutions made in this time. Ans~N~-~ 5 rad/sz, 20.4 rev
9.24
A car wheel 30 cr~n radius is turning at a rate of 8 rev/s when the car begin~.o slow uniformly to rest in
a time of 14 ~nd the number of revolutions made by the wheel and the die, nee the car goes in the
14s. Ay 56 rev, 106.m
9.25
A wheel~evolving at 6 rev/s has an angular acceleration of 4 rad/s2. Find the number b~turns the wheel
mus~ake toacquire an angular speed of 26 re v/s and the time required. Arts. 50 e2~, 31.4s
/
\
/
9.2g ~(~tring wound on the rim of a wheel 20 cm in diameter is pulled out at a rate of 75 cm/~ Thr’o~gh how
/many revolutions will the wheel have turned by the time that 9.0m of string has been unwound~How
4’ long will it take? Ans. 14.3 rev, 12 s
"~
A mass of 1.5
kg
moves ina circle of radius 25cm at 2rev/s. Calculate (a) the tangential velocity, (b)
¯o
the acceleration, and (c) the required centripetal force for the motion, it
V’~a.~\~ ~c ~2~
Ans. (a) 3.14m/s; (b) 39.4m/s2 radially inward; (c) 59N
(a) Compute the radial acceleration of a point the equator of the earth. (b) Repeat for the north pole
of the earth. Take the radius of the earth to be 6370 kin. Am. (a) 0.0337 m/sZ; (b) zero
A car moving at 5 m/s tries to round a corner in a circular arc of 8 m radius. The roadway is flat¯ How
large must the coefficient of friction be between wheels and roadway if the car is not to
skid? Arts. 0.32
A box rests at a point 2.0 m from the axis of a horizontal circular platform. The coe~cient of static
friction between box and platform is 0.25. As the rate of rotation of the platform is slowly increased
from zero, at what angular speed will the box first slide? Ans¯ 1,107 rad/s
A stone rests in a pail that is moved in a vertical circle of radius 60 cm. What is the least speed the stone
must have as it rounds the top of the circle if it is to remain in contact with the pail? Ans. 2.4 m/s
94
ANGULAR MOTION IN A PLANE
[CHAP. 9
A pendulum 80 cm long is pulled to the side, so that its bob is raised 20 cm from its lowest position, and
is then r, eleased. As the 50 g bob moves through its lowest position, (a) what is its speed and (b) what is
the tension in the pendulum cord? Arts. (a) 1.98 m/s; (b) 0.735 N
Refer back to Fig. 9-4. How large must h be (in terms of R) if the frictionless wire is to exert no force on
the bead as it passes through point B? Assume the b~ad is released from rest at A. Ans. 2.5R
9~
If, in Fig. 9-4 and in Problem 9.33, h = 2.5R, how large a force will the 50 g bead exert on the wire as it
passes through point C? Ans. 2.94N
radius 6570kin. Find the speed of the
(~.3~ A satellite orbits the earth at a height of 200 km in a circle of
satellite and the time taken to complete one revolution. Assume the earth’s mass is 6.0 x I024 kg. (Hint:
~
The gravitational force provides the centripetal force.) Arts. 7.8 kin/s, 88 min
A roller coaster is just barely moving as it goes over the top of one hill. It rolls nearly without friction
down the hill and then up over a lower hill that has a radius of curvature of 15 m. How much higher
must the first hill be than the second if the .passengers are to exert no force on the seat as they pass over
the top of the lower hill? Arts. 7.5 m
The human body can safely stand an acceleration 9 times that due to gravity. With what minimuur radius
of curvature may a pilot safely turn the plane upward at the end of a dive if the plane’s speed is
770kin/h? Arts. 519 m
A 60kg glider pilot traveling in a glider at 40m/s wishes to turn an inside vertical loop such that he
exerts a force of 350 N on the seat when the glider is at the top of the loop. What must be the radius of
the loop under these conditions? (Hint: Gravity and the seat exert forces on the pilot.) Arts. 102 m
Suppose- the earth is a perfect sphere with R = 6370 kin. Ira person weighs exactly 600.0 N at the north
pole, how much will the person weigh at the equator? (Hint: The upward push of the scale on the
person is what the scale will read and is what we are calling the weight in this case.) Ans~ 597.9 N
A mass m hangs at the end of a pendulum of length L which is released at an angle of 40° to the vertical.
Find the tension in the pendulum cord wheu it makes an angle of 20° to the vertical (Hint: Resolve the
weight along and perpendicular to the cord.) Ans. 1.29mg