Physics 53
Summer 2009
Exam I
Solutions
Part A: Multiple choice questions. Check the best answer.
1.
The total mechanical energy of a particle:
Always decreases if non-conservative forces do work. [Can increase]
Is always positive.
Changes by the total amount of work done by all forces. [Only nonconservative forces]
√
2.
Is constant if only conservative forces do work.
Shown are trajectories for three projectiles launched with the same initial speed.
3
2
1
Which of the following is NOT true?
#1 reaches its target in the least time.
#3 stays in flight the longest.
The elevation angles at launch for #1 and #2 add to 90°.
√
A projectile launched at an angle between those of #1 and #2 will not go
as far as either of those. [Farther]
1
Physics 53
3.
Summer 2009
In the situation shown, a block is being dragged across a rough floor by pulling
on the rope with force T. The coefficient of kinetic friction is µ k . The force of
friction on the block has magnitude:
µk mg
T
µkT sin !
√
θ
µk (mg ! T sin " )
Cannot be determined without knowing the block’s acceleration.
4.
The total potential energy for a combination of conservative forces is given by
U(x) = x 2 ! 2x . Which of the following is NOT true?
The system has only one point of static equilibrium, at x = 1 .
The formula for the total force is Fx = 2 ! 2x .
√
The equilibrium at x = 1 is unstable. [Stable]
If the total energy is zero, there are turning points at x = 0 and x = 2 .
Part B: True-false questions. Mark each sentence with T of F. Each question has a value of 3 points.
1.
The purpose of a simple machine (such as an inclined plane or a lever) is to do
the same work with a smaller force over a larger distance.
T
2.
A bucket of water is sliding across a floor, and the
surface of the water is tilted as shown; the bucket is
accelerating to the right.
F
3.
[Surface is perpendicular to g eff , so acceleration is to the left]
Pushing a heavy box at constant speed up an frictionless incline requires the
same power input as lifting it straight up at the same speed to the same height.
F
[Same energy input, but less power over longer time]
2
Physics 53
Summer 2009
Part C: Problems. Work problems in the space provided, indicating your method clearly. A correct final
answer supported by no argument, or a fallacious one, will receive little or no credit. The problem carry
the point values shown.
1.
A stone is propelled from a catapult on the ground, with these initial velocity
components: 40 m/s horizontal, 30 m/s vertical. Use g = 10 m/s2 .
a.
How long will it be in flight?
b.
How high will it rise?
c.
How far from the launch site will the stone land on the ground again?
d.
An archer standing on the ground under the point at which the stone
reaches its highest point fires an arrow straight up, trying to hit the stone.
He shoots his arrow 2 s after the stone is launched. What must be the
initial speed of the arrow if it is to hit the stone?
[20 points]
a.
Vertical velocity: v y = 30 ! gt = 30 ! 10t . This is zero when t = 3 s, so the
total time of flight is twice this, or 6 s.
b.
Vertical position: y = 30t ! 12 gt 2 = 30t ! 5t 2 . With t = 3 s we find ymax = 45
m.
c.
Horizontal position: x = 40t . With t = 6 s we find xmax = 240 m.
d.
It must rise 45 m in 1 s, so we have 45 = v0 (1) ! 12 (10)(1)2 , or v0 = 50 m/s.
3
Physics 53
2.
Summer 2009
An enclosed car moves without friction on a track, starting with speed v0 at the
position shown. When it reaches point A its speed is such that the track exerts no
force on it. The radius of curvature of the track at both points A and B is h/2.
A
h
B
a.
What is v0 ?
b.
A mass m attached to a massless string is hanging from
the ceiling of the car as shown in the picture. What is
the tension in the string when the car is at point A?
c.
What is the tension in the string when the car is at point B?
[Give all answers in terms of m, g and h.]
[15 points]
a.
If the track exerts no force at the top, the only force is gravity, so the
downward acceleration is g. The radial acceleration is vA 2 /(h/ 2) = g , so
vA 2 = gh/ 2 . Use conservation of energy, with U = 0 at the bottom of the
track. We have
1
mv0 2
2
= 12 mvA 2 + mgh , or v0 2 = vA 2 + 2gh = 52 gh .
b.
Since the car’s acceleration at A is g, we have g eff = g ! g = 0 , so the
tension in the string is zero. (Everything in the car is weightless.)
c.
At B the speed is again v0 , so the (upward) radial acceleration must be
ar = v0 2 /(h/ 2) = 5g . The forces on the mass are its weight (down) and the
tension (up), so we have T ! mg = mar = 5mg , or T = 6mg .
4
Physics 53
3.
Summer 2009
In a test of tires a car travels a flat circular track of radius 32 m in time 4π s at
constant speed v.
a.
What is the radial acceleration (magnitude)?
b.
What is the minimum coefficient of static friction between tires and road?
c.
With the same tires and road, what is the shortest distance the car can stop
from initial speed v if it moves in a straight line?
d.
If the circular track is banked so as not to require friction at speed v, what
is the tangent of the banking angle? [Draw the free body diagram.]
[All answers are numerical. Use g = 10 m/s2 .]
[20 points]
a.
The speed is v = (2! " 32)/ 4! = 16 m/s. The radial acceleration is
ar = v 2 /R = (16)2 /32 = 8 m/s2 .
b.
Static friction must provide the acceleration, so fs (max) = µs mg = mv 2 /R .
Thus µs = v 2 /Rg = 0.8 .
c.
Use W f = !E . We have d = ! µs mg " d = 0 ! 12 mv 2 , or d = v 2 / 2 µs g = 16 m.
[Note that you can stop in a straight line distance equal to half the radius
of the smallest circle you can turn in.]
d.
Diagram below. Horizontal forces: N sin ! = mar . Vertical forces:
N cos! = mg . Solve for the angle by dividing: tan ! = ar / g = 0.8 .
N
θ
5
mg
Physics 53
4.
Summer 2009
A block of mass m is resting on an incline, at height h above the bottom. The
incline has sin ! = 3/5 , cos! = 4 /5 . The static friction coefficient is such that the
block is on the verge of slipping.
k
h
θ
a.
What is µs ? [Numerical answer]
b.
The block is given a nudge to start it sliding, with µ k = 32 µs . What is the
work done by friction as it slides to the bottom, in terms of m, g, and h.
c.
It now slides across a frictionless horizontal surface until it hits a spring
with stiffness k, compressing it distance x before coming momentarily to
rest. What is x, in terms of the given quantities? [What is the block’s
kinetic energy at the bottom of the incline?]
d.
The block then slides back to the left and up the incline until it comes to
rest. At what height h’ does it stop? [Answer in terms of h only.]
[20 points]
a.
Forces down the incline: mg sin ! " fs = 0 . Forces normal to incline:
N ! mg cos" = 0 . Maximum friction fs = µs N : Solving we find
µs = tan ! = 3/ 4 .
b.
The kinetic friction force is f k = µ k N = (1/ 2)mg cos! = (2/5)mg . The
distance the block moves on the incline is d = h/sin ! = (5/3)h . The work
by friction is thus W f = ! f k d = !(2/3)mgh .
c.
The kinetic energy at the bottom is K = mgh ! (2/3)mgh = (1/3)mgh . This
2gh
is converted into spring potential energy, so 12 kx 2 = 13 mgh , or x 2 =
.
3k
d.
By the same steps as in (b) we find that the work by friction as the block
rises to height h’ is W f ' = !(2/3)mgh' . The final total energy is mgh' , so
we have mgh'! (1/3)mgh = !(2/3)mgh' . Solving we find h' = h/5 .
6
Physics 53
Summer 2009
Median = 70
SD = 16.3
7