Chapter 4
Motion With a Changing Velocity
P2.27: Find the magnitude and
direction of the vector with the
following components:
(a) x = -5.0 cm, y = +8.0 cm
(b) Fx = +120 N, Fy = -60.0 N
(c) vx = -13.7 m/s, vy = -8.8 m/s
(d) ax = 2.3 m/s2, ay = 6.5 cm/s2
P2.56: A 3.0 kg block is at rest on a
horizontal floor. If you push
horizontally on the block with a force of
12.0 N, it just starts to move.
(a) What is the coefficient of static
friction?
(b) A 7.0-kg block is stacked on top of the
3.0-kg block. What is the magnitude F
of the force acting horizontally on the
3.0-kg block as before, that is required
to make the two blocks start to move?
P3.47: A 2010-kg elevator moves with
an upward acceleration of 1.50 m/s2.
What is the tension that supports
the elevator?
P3.48: A 2010-kg elevator moves with
a downward acceleration of 1.50
m/s2. What is the tension that
supports the elevator?
Kinematic Equations for Const. Acceleration
Fnet = ma.
If Fnet is const, a will also be const.
Uniformly accelerated motion: a = const.
A car moves at a constant acceleration of
magnitude 5 m/s2. At time t = 0, the
magnitude of its velocity is 8 m/s. What is
the magnitude of its velocity at
(i) t = 2s? (ii) t = 4s? (iii) t = 10s?
A car moves at a constant acceleration of
magnitude 5.7 m/s2. At time t = 0, the
magnitude of its velocity is 18.3 m/s.
What is the magnitude of its velocity at
t = 2.2s?
Kinematic Equations for Const. Acceleration
Consider an object on which a net force
Fnet acts on it. Thus it moves with an
acceleration.
As the object moves, its velocity changes.
Fnet
a
Time = 0
Initial position = x0
Initial velocity = v0
Fnet
Fnet
a
Time = t
Final position = x
Final velocity = v
Kinematic Equations for Const. Acceleration
Fnet = ma.
If Fnet is const, a will also be const.
Uniformly accelerated motion:
a = const.
• aave = ainst
• Let us use initial time, t1 = 0.
•Final time, t2 = t, hence t = t – 0 = t
• Position: initial, x1 = x0, final, x2 = x
• Velocity, initial v1 = v0, final, v2 = v
Kinematic Equations for Constant Acceleration
Uniformly accelerated motion: a = constant.
Time: Initial = 0, final = t
Positions: Initial = x0, final = x
Velocity: Initial = v0, final = v
v = v0 + at
x = x0 +vot + ½ at2
v2 = v02 + 2a(x-x0)
Average velocity vav = (v0 + v)/2
Example Problem 4.14
A train traveling at a constant speed of
22 m/s, comes to an incline with a constant
slope. While going up the incline the
train slows down with a constant
acceleration of magnitude 1.4 m/s2.
(a)Draw a graph of vx versus t.
(b)What is the speed of the train after 8.0s
on the incline?
(c) How far has the train traveled up the
incline after 8.0 s?
A car moving south slows down with at a constant
acceleration of 3.0 m/s2. At t = 0, its velocity is 26
m/s. What is its velocity at t = 3 s?
A.
B.
C.
D.
E.
35 m/s south
17 m/s south
23 m/s south
29 m/s south
17 m/s north
1
2
3
4
5
6
7
8
9
10
11
12
21
22
23
24
25
26
27
28
29
30
31
41
42
43
44
45
46
47
48
49
50
51
61
62
63
64
65
14
15
16
17
18
19
20
32
0% 33
34
0%
35
36
0%
37
38
0%
39
40
0%
52
54
55
56
57
58
59
60
A.
13
53
B.
C.
D.
E.
A car initially traveling at 18.6 m/s begins to slow
down with a uniform acceleration of 3.00 m/s2. How
long will it take to come to a stop?
A.
B.
C.
D.
E.
55.8 s
15.6 s
6.20 s
221.6 s
None of these
1
2
3
4
5
6
7
8
9
10
11
12
21
22
23
24
25
26
27
28
29
30
31
41
42
43
44
45
46
47
48
49
50
51
61
62
63
64
65
14
15
16
17
18
19
20
32
0% 33
34
0%
35
36
0%
37
38
0%
39
40
0%
52
54
55
56
57
58
59
60
A.
13
53
B.
C.
D.
E.
Free Fall
• Free fall: Only force of gravity acting on
an object making it fall.
• Effect of air resistance is assumed
negligible.
• Force of gravity acting on an object near
the surface of the earth is F = W = mg.
• Acceleration of any object in free fall:
a = g = 9.8 m/s2 down (ay = -9.8 m/s2).
Free Fall
+y
ay = -9.8 m/s2
ax = 0
+x
Free Fall contd…
1. a = g, regardless of mass of object.
2. a = g, regardless of initial velocity
ay = -9.8 m/s2, ax = 0
+y
v0 = 0
v0 = -15 m/s
v0 = +15 m/s
+x
3. Free Fall: Motion is symmetric.
ay = -9.8 m/s2, ax = 0
+y
At the maximum height:
• vy = 0
• Speed at equal heights will be
equal.
• Equal time going up and down.
v0 = +5 m/s
+x
Example: Problem 4.32
A stone is launched straight up by a
slingshot. Its initial speed is 19.6 m/s
and the stone is 1.5 m above the
ground when launched.
(a) How high above the ground does the
stone rise?
(b) How much time elapses before the
stone hits the ground?
APPARENT WEIGHT
A physics student whose mass is 40 kg
stands inside an elevator on a scale that
reads his weight in Newtons.
Scale Reading = Normal force the scale
exerts on the student.
N
mg
Scale Reading = N
= mg
= 40 x 9.8 N
= 392 N
1. Elevator at rest. What will be the scale
reading?
N
Fnet = N – W = may
W = mg
At rest means ay = 0.
Hence N = W, ie apparent weight = true
weight = 40 x 9.8 = 392 N
2. Elevator accelerating upwards with ay = 2.0
m/s2. What will be the scale reading?
2. Elevator accelerating upwards with ay =
2.0 m/s2. What will be the scale reading?
N
Fnet = N – W = may
ay = + 2.0 m/s2 (positive because
a
acceleration is upwards) . Hence,
W = mg
N –W = N – mg = may.
N = mg + may = m(g+ay) = 40(9.8 + 2.0)
= 472 N
ie, apparent weight is greater than the true
weight.
3. Elevator accelerating downwards with ay
= 2.0 m/s2. What will be the scale reading?
3. Elevator accelerating downwards with ay
= 2.0 m/s2. What will be the scale reading?
N
Fnet = N – W = may
a
ay = - 2.0 m/s2 (negative because
acceleration is downwards) .
W = mg
Hence N –W = N – mg = -may.
N = mg - may = m(g - ay) = 40(9.8 - 2.0) =
312 N
ie, apparent weight is less than the true
weight.
A 112.0-kg person stands on a scale inside an elevator
moving downward with an acceleration of 1.80 m/s2.
What will be the scale reading?
A.
B.
C.
D.
E.
81%
1299 N
1,098 N
896 N
112 N
0N
12%
7%
0%
A.
B.
C.
D.
0%
E.
A ball is kicked straight up from ground level with
initial velocity of 22.6 m/s. How high above the
ground will the ball rise?
73%
A.
B.
C.
D.
E.
9.8 m
3.00 m
1.15 m
26.1 m
19.6 m
13%
5%
A.
8%
3%
B.
C.
D.
E.
WEIGHTLESSNESS
If the elevator was going down with an
acceleration ay = g = -9.8m/s2, then
N = m(g-g) = 0
ie, apparent weight = 0
This is “weightlessness” or “zero gravity”
Apparent weight of an object in free fall is
zero while its true weight remains
unchanged.
Equilibrium
Newton’s 2nd Law: Fnet = ma
For an object in equilibrium: Fnet = 0
Static (v = 0) 0r dynamic (v = constant) eqlbm.
2-dimensions, separate the x and y components and
treat the problem as two 1-dim problems.
Fx = max
Fy = may
For equilibrium, Fx = max = 0 and  Fy = may = 0
y
2-Dimensions
• X and Y are INDEPENDENT!
• Break 2-D problem into two
1-D problems.
x
Equilibrium
Determine the tension in the
6 m rope if it sags 0.12 m in
the center when a gymnast
with weight 250 N is standing
on it.
x direction: Fx = max = 0
-TL cosq + TR cosq = 0
TL = TR
y direction: Fy = may = 0
TL sinq + TR sinq - W = 0
2 T sinq = W
T = W/(2 sinq) = 3115 N
y
TR
TL
x
W
q
3m
0.12
tan q =
3
TR
.12 m
q = 2.3
Equilibrium on a Horizontal Plane
Object at rest or moving with const. velocity
Fx = max = 0 and  Fy = may = 0
Object at rest
N
fs
F
W
No motion until F = > fsmax
Fx = F - fs = 0 or F = fs.
Fy = N - W = 0 or N = W
Sliding with constant velocity
N
fk
F
W
Fx = F - fk = 0 or F = fk
Fy = N - W = 0 or N = W
Equilibrium on an inclined Plane
An object at rest on an inclined plane
Fx = max = 0 and  Fy = may = 0
x
x
q
W
q
q
W
W
Wx
x
q
Equilibrium on an inclined Plane
An object at rest on an inclined plane
Fx = max = 0 and Fy = may = 0
x
q
Fy = may = 0
N - Wcosq = 0 or N = Wcosq
Fx = max = 0
fs - Wsinq = 0 or fs = Wsinq
If angle q is increased, the object will eventually slide
down the plane. Sliding will start beyond angle qmax
At qmax: fsmax = W.sinqmax. But fsmax = sN = s(Wcosqmax)
Therefore, sWcosqmax = Wsin qmax
OR s = (Wsinqmax)/ Wcosqmax ie, s = tanqmax
Equilibrium on an inclined Plane
An object at rest on an inclined plane
x
q
N = Wcosq
fs = Wsinq
• If angle q is increased, the object will eventually
slide down the plane.
• Sliding will start beyond angle qmax
• At qmax: fsmax = W.sinqmax.
• But fsmax = sN = s(Wcosqmax)
• Therefore, sWcosqmax = Wsin qmax
• OR s = (Wsinqmax)/ Wcosqmax ie, s = tanqmax
A mass m being pulled uphill by a force F
x If m = 510 kg, s= 0.42, k =
q
0.33, q = 15o:
(a) Find minimum force F
needed to start the mass
moving up.
(b)If the force in (a) is
maintained on the mass,
what will its acceleration
be?
(c) To move the mass with constant speed, what must
the value of F be?
A block is at rest on a flat board. The flat board is
gently tilted. At what angle will the block start to
slide? Assume the coefficient of static friction (s)
between the block and the board is 0.48.
95%
A.
B.
C.
D.
E.
0.48o
61.3o
28.7o
25.6o
0.00837o
2%
A.
0%
B.
2%
0%
C.
D.
E.
Position, Velocity and Acceleration
• Position, Velocity and Acceleration are
Vectors! x direction y direction
 
rf  r0

vav =
t f  t0
x f  x0
vx =
t f  t0
y f  y0 
2
2
vy =
v = vx  v y
t f  t0
 
v f  v0

aav =
t f  t0
vxf  vx 0
ax =
t f  t0
v yf  v y 0 
2
2
a = ax  a y
ay =
t f  t0
• x and y directions are INDEPENDENT!
Velocity in Two Dimensions
A ball is rolling on a horizontal surface at 5 m/s. It then
rolls up a ramp at a 25 degree angle. After 0.5 seconds,
the ball has slowed to 3 m/s. What is the change in
velocity?
x-direction
y-direction
vix = 5 m/s
viy = 0 m/s
vfx = 3 m/s cos(25)
vfy = 3 m/s sin(25)
vx = 3cos(25)–5 =-2.28m/s
vy = 3sin(25)=+1.27 m/s
y
x
v = v  v = 2.6 m/s
2
x
2
y
3 m/s
5 m/s
Acceleration in Two Dimensions
A ball is rolling on a horizontal surface at 5 m/s. It then
rolls up a ramp at a 25 degree angle. After 0.5 seconds,
the ball has slowed to 3 m/s. What is the average
acceleration? [Assume force of gravity is very small].
y
x-direction
y-direction
 2.28m/s
ax =
= 4.56 m/s 2
0.5 s
1.27m/s
ax =
= 2.54 m/s 2
0.5 s
a = a  a = 5.21 m/s 2
2
x
2
y
3 m/s
5 m/s
x
A wagon of mass 50 kg is being pulled
by a force F of magnitude 100 N
applied through the handle at 30o
from the horizontal. Ignoring
friction, find the magnitude of
(a) the horizontal component of F.
(b) the horizontal component of
acceleration.
(c) the normal force exerted on the
wagon.
Projectile Motion
A projectile – An object moving in 2dimensions near the surface of the earth with
only the force of gravity acting on it.
Eg: golf ball, batted base ball, kicked
football, soccer ball, bullet, etc.
• Assume no air resistance.
• Assume g = -9.8 m/s2 constant.
• We are not concerned with the process that
started the motion!
Free Fall: 1-dimensional motion.
ay = -9.8 m/s2, ax = 0
+y
At the maximum height:
• vy = 0
• Speeds at equal heights will
be equal.
• Equal time going up/down.
v0 = +5 m/s
+x
PROJECTILE: Free Fall motion in 2-dimensions.
ay = -9.8 m/s2,
ax = 0
+y
v0 = 5 m/s
q
v0x
v0y
+x
PROJECTILE: Free Fall motion in 2-dimensions.
ay = -9.8 m/s2
ax = 0
v0x = v0cosq
v0y = v0sinq
What will happen to the y-component of the
velocity?
What will happen to the x-component of the
velocity?
Kinematics in Two Dimensions
• x = x0 + v0xt + ½ axt2
• y = y0 + v0yt + ½ ayt2
• vx = v0x + axt
• vy = v0y + ayt
• vx2 = v0x2 + 2ax (x - x0)
• vy2 = v0y2 + 2ay (y – y0)
x and y motions are independent!
They share a common time t.
Kinematics for Projectile
Motion
ax = 0
ay = -g
• x = x0 + v0t
• y = y0 + v0yt - 1/2 gt2
• vx = v0x
• vy = v0y - gt
X
• vy2 = v0y2 - 2g y
Y
PROJECTILE: Free Fall motion in 2-dimensions.
ay = -9.8 m/s2, ax = 0
Once the projectile is in air, the only
force acting on it is gravity. Its trajectory
(path of motion) is a parabola.
Fnet = ma = -mg
ay = -9.8 m/s2
ax = 0
PROJECTILE: Free Fall motion in 2-dimensions.
ay = -9.8 m/s2 and ax = 0
v0x = v0cosq and v0y = v0sinq
Two balls A and B of equal mass m. Ball A is
released to fall straight down from a height h. Ball B
is thrown horizontally. Which ball lands first?
A
B
h
ay = -9.8 m/s2
ax = 0
Vo = 0
v0x = 0
v0y = 0
ay = -9.8 m/s2
ax = 0
Vo 0
v0x = Vo
v0y = 0
A
ay = -9.8 m/s2
ax = 0
vo = 0
v0x = 0
v0y = 0
y0 = 0, y = -h
• y = y0 + v0yt + ½ ayt2
• vy = v0y + ayt
• vy2 = v0y2 + 2ay (y – y0)
To find time t, use
y = y0 + v0yt + ½ ayt2
-h = 0 + (0 . t) + ½ (-g)t2
Gives 2h = gt2 and t = (2h/g)
ay = -9.8 m/s2
ax = 0
Vo 0
v0x = vo
v0y = 0
B
• y = y0 + v0yt + ½ ayt2
• vy = v0y + ayt
• vy2 = v0y2 + 2ay (y – y0)
To find time t, use
y = y0 + v0yt + ½ ayt2
-h = 0 + (0 . t) + ½ (-g)t2
Gives 2h = gt2 and t = (2h/g)
A flatbed railroad car is moving along
a track at constant velocity. A
passenger at the center of the car
throws a ball straight up. Neglecting
air resistance, where will the ball
land?
1. Forward of the center of the car
2. At the center of the car
3. Backward of the center of the car
Since no air resistance is present, the ball
and the train would be moving with the
same horizontal velocity, and when the ball
is tossed, it is given an additional velocity
component in the vertical direction, but
the original horizontal velocity component
remains unchanged, and lands in the center
of the train.
P 4.22
A penny is dropped from the
observation deck of the Empire
State building (369 m above the
ground). With what velocity does it
strike the ground? Ignore air
resistance.
P 4.36
An arrow is shot into the air at an
angle of 60.0o above the horizontal
with a speed of 20.0 m/s.
(a) What are the x- and y-components
of the velocity of the arrow 3.0 s
after it leaves the bowstring?
(b) What are the x- and y- components
of the displacement of the arrow
during the 3.0-s interval?
A ball is thrown with a speed of 40.0 m/s at 55o
above the horizontal. At the maximum height, its
speed will be
47%
A.
B.
C.
D.
E.
22.9 m/s
-9.8 m/s
0 m/s
32.8 m/s
40.0 m/s
30%
14%
9%
0%
A.
B.
C.
D.
E.
A ball is thrown with a speed of 40.0 m/s at 35o
above the horizontal. How long is it in air?
25%
25%
23%
A.
B.
C.
D.
E.
6.69 s
8.16 s
2.34 s
4.08 s
4.68 s
20%
8%
A.
B.
C.
D.
E.
A ball is kicked straight up from ground level with
initial velocity of 22.6 m/s. How high above the
ground will the ball rise?
69%
A.
B.
C.
D.
E.
9.8 m
3.00 m
1.15 m
26.1 m
19.6 m
8%
8%
A.
B.
10%
5%
C.
D.
E.
Summary of Concepts
• X and Y directions are Independent!
– Position, velocity and acceleration are vectors
• F = m a applies in both x and y direction
• Projective Motion
– ax = 0 in horizontal direction
– ay = g in vertical direction
50
1. A car initially traveling at a velocity vo begins to
slow down with a uniform deceleration of 1.20 m/s2
and comes to a stop in 26.0 seconds. Determine the
value of vo.
95%
A.
B.
C.
D.
E.
31.2 m/s
21.7 m/s
27.2 m/s
24.8 m/s
None of these
5%
A.
B.
0%
0%
C.
D.
0%
E.
2. A 102.0-kg person stands on a scale inside an
elevator moving downward with an
acceleration of 1.300 m/s2. What will be his
apparent weight?
91%
A.
B.
C.
D.
E.
999.6 N
132.6 N
867.0 N
1132 N
0N
7%
2%
0%
A.
B.
C.
D.
0%
E.
3. A ball is thrown with a speed of 27.0 m/s at 35o
above the horizontal. At the maximum height, its
speed will be
82%
A.
B.
C.
D.
E.
27.0 m/s
-9.8 m/s
0 m/s
22.1 m/s
15.5 m/s
16%
2%
A.
0%
B.
0%
C.
D.
E.
4. A ball is thrown with a speed of 32.0 m/s at 50o
above the horizontal. How long is it in air?
64%
A.
B.
C.
D.
E.
4.20 s
6.53 s
3.27 s
2.50 s
5.00 s
11%
13%
7%
A.
B.
C.
4%
D.
E.
5. A ball is kicked straight up from ground level with
initial velocity of 16.6 m/s. How high above the
ground will the ball rise?
93%
A.
B.
C.
D.
E.
28.1 m
14.1 m
1.69 m
0.847 m
1.18 m
2%
A.
B.
2%
2%
C.
D.
0%
E.
6. A block is at rest on a flat board. The flat board
is then gently tilted. If the block starts to slide at a
tilt angle of 23.8o, what is the coefficient of static
friction (s) between the block and the board? 89%
A.
B.
C.
D.
E.
0.40
0.91
23.8
87.6
0.44
9%
0%
0%
A.
B.
2%
C.
D.
E.