QUIZ 4
Physics 101, Fall 2008
October 10, 2008
SOLUTIONS
Problem 1 [2 pts]. At the bowling alley, the ball-feeder mechanism must exert a steady force to
push the bowling balls up a 1.0-m long ramp. The balls are sliding without rolling along the ramp,
which leads the balls to a chute 0.5 m above the base of the ramp. Sliding friction is negligible.
Approximately how much force must be exerted on a 5.0-kg bowling ball?
L
H
•
•
•
•
•
200 N
50 N
25 N
5.0 N
impossible to determine
The force exerted by the mechanism times the distance of 1.0-m over which the force is exerted must
equal the change in the potential energy of the ball: mgH = F L. It follows that:
F =
5.0 kg · 9.8 ms−2 0.5 m
mgH
=
= 24.5 N
L
1m
The correct answer among the choices presented is 25 N .
Problem 2 [2 pts]. A sports car accelerates from zero to 40 km/h in 1.5 s. How long does it take
for it to accelerate from zero to 80 km/h, assuming the power of the engine to be independent of
velocity and neglecting friction?
•
•
•
•
•
•
2s
3s
4.5 s
6s
9s
12 s
Answer: 6 s. In the absence of friction, the power of the engine is equal to the kinetic energy of the car
divided by the time it took to attain that kinetic energy.
QUIZ 4
Physics 101, Fall 2008
October 10, 2008
Problem 3. Two blocks of mass M (25.5 kg) and m (17.3 kg) are suspended on the pulley system
as shown, and initially at rest. Note that the block with mass M is suspended while the block with
mass m is resting on an inclined surface. The coefficient of kinetic friction µk between the inclined
surface and the block with mass m is 0.45. Other data: h=15 m, θ=55◦ .
M
h
μK
m
θ
a. [1 pts] Draw a diagram of all forces acting on the two bodies.
T
Mg
N
μ KN
mg
b. [2 pts] What is the work performed by friction on the block of mass m while the block of mass
M falls from its initial position to the horizontal surface below it?
Since the block of mass m does not accelerate in the direction perpendicular to the incline, the normal
force must balance the component of the weight force in the direction perpendicular to the incline:
N = mg cos θ.
While the block of mass M falls by a distance h, the block of mass m moves by the same distance h
along teh incline and rises by the distance h sin θ.
The work performed by the friction:
Wnc = −µk N h = −µK mg cos θh = −0.45 · 17.3 kg · 9.8 ms−2 · cos 55◦ · 15 m = -656 J .
c. [3 pts] What is the velocity vf of the block with mass M just before touching the horizontal
surface?
1
Wnc = ∆KE + ∆P E = (m + M )vf2 − M gh + mgh sin θ
2
r
2
(Wnc + M gh + mgh sin θ)
vf =
m+M
r
vf =
2
(−656 J + 25.5 kg · 9.8 ms−2 · 15 m − 17.3 kg · 9.8 ms−2 · 15 m · sin 55◦ ) = 6.49 m/s.
42.8 kg