FULL ANSWER PROBLEM (6 points): Answers to the following questions should be inserted
in the spaces provided. Show all works in spaces provided. DO N OT FORGET UNITS!
Consider the Atwood’s machine where m1 = 5 kg and m2 = 8 kg. The pulley
can be considered to be frictionless and the rope massless.
(I)
(2 points) Draw the free body diagrams for m1 and m2 separately.
(II)
(2 points) Write Newton’s second law for m1 and m2 separately.
T – m1g = m1 a
m2g – T = m2 a
(III)
(2 points) Solve the above equations to find the magnitude of the acceleration of the
blocks and the magnitude of the tension in the rope.
𝑚𝑚2 − 𝑚𝑚1
8−5
𝑚𝑚
𝑎𝑎 = �
� 𝑔𝑔 = �
� 9.81 = 2.26 2
𝑚𝑚1 + 𝑚𝑚2
5+8
𝑠𝑠
T = m1(g+a) = 5(9.81 + 2.26) = 60.4 N
2
Multiple Choice Questions
1. From the top of a building, a stone is thrown with an initial speed at an angle of 60°
above the horizontal and hits the ground 3.0 s later. The horizontal distance from the
building to the point where the stone strikes the ground is 12 m. Find the vertical height
of the building.
(A) 23 m
(B) 31 m
(C) 57 m
(D) 65 m
(E) none of these
12 = v cos60 *3, v= 8 m/s
y = v sin60 t – gt2/2 = 8 sin 60 *3 -4.9*32 = -23.3 m
2. At a given instant the train engine at E has a speed of 18 m/s and a total acceleration of 20 m/s2
which makes 75 degree angle with the velocity. Determine the radius ρ of the path.
(A) 17.5 m
(B) 16.2 m
(C) 16.8. m
ar = v2/r = a total sin 75,
r = v2/ (a total sin 75) = 182/
(20 sin 75) = 16.77 m
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(D) 29.6 m
(E) 62.6 m
3. Suppose an object is moving in circular motion in the 𝑥𝑥𝑥𝑥 plane. At a given instant, the
m
m
m
velocity vector is 𝐯𝐯 = −5 s 𝐣𝐣̂ and the acceleration is 𝐚𝐚 = 10 s2 𝐢𝐢̂ − 15 s2 𝐣𝐣̂. Which of the
following is true?
a. The object is moving counter-clockwise around the circle and is slowing down
b. The object is moving counter-clockwise around the circle and is speeding up
c. The object is moving clockwise around the circle and is slowing down
d. The object is moving clockwise around the circle and is speeding up
𝑦𝑦
𝑥𝑥
4. A pilot executes a uniform circular path in an airplane. The initial velocity of the plane is
given as v0 = {2500i + 3000j} m/s. One minute later, the velocity of the plane is v = {2500i – 3000j} m/s. Find the magnitude of the acceleration at this later point in time.
a) 119 m/s2
b) 204 m/s2
c) 262 m/s2
d) 314 m/s2
e) 409 m/s2
SOLUTION
Based on the information the period of the plane is T = 120 s = 2 min. The magnitude of the
velocity is,
v = (25002 + 30002)1/2 = 3905 m/s
a = v2/r and since T = 2πr/v then,
a = 2πv/T = 2π(3905)/120 = 204 m/s2
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5. In the figure the force of static friction is keeping the blocks at rest
with zero acceleration. Which of the following is true of the force
of static friction that block 2 exerts on block 1?
a. It points up
b. It points down
c. It is zero
d. Whether or not it points up or down depends on the masses of the
two blocks and the applied force
wall
1
2
6. A 1g insect can jump straight up at 4 m/s2. The contact with the ground lasts 1 ms before
the insect is clear of the ground. Which of the following statements is true?
a) During the jump the normal force of the ground on the insect is greater than the force
of gravity on the insect.
b) During the jump the normal force of the ground on the insect is less than the force of
gravity on the insect.
c) During the jump the normal force of the ground on the insect is equal to the force of
gravity on the insect?
Solution
(a) While the insect is in the act of jumping, it experiences an upward acceleration of 4
m/s 2 , so the net force acting on it must be upward. Because only the normal force and the
force due to gravity are acting in the vertical direction, the normal force from the ground
must be greater than the force due to gravity.
7. The horizontal surface on which the block slides is frictionless. If F = 20 N and M = 5.0
kg, what is the magnitude of the resulting acceleration of the block?
a. 5.3 m/s2
b. 6.2 m/s2
c. 7.5 m/s2
d. 4.7 m/s2
5
𝑭𝑭
e. 3.2 m/s2
8. Suppose a chair with wheels sits on a rough carpet. You apply a force to the chair to bring
it from rest to a constant velocity and then let go of the chair. The chair then travels some
distance before coming to rest. Which of the following statements correctly explains why
the chair comes to rest after you release it?
a. The chair comes to rest because forces are required to keep objects moving.
b. The chair comes to rest because its inertia (i.e. mass) slows it down.
c. The chair comes to rest because there is a force causing it to slow down.
d. The chair comes to rest because all objects must eventually slow down.
9. What is the maximum speed with which a 1500 kg car can make a right turn around a
curve of radius 50 m on a level (un-banked) road without sliding? The coefficient of
kinetic friction between the car and the road is 0.10. The coefficient of static friction is
0.40.
(A) 15.7 m/s
(B) 49.0 m/s
(C) 14.0 m/s
(D) 7.00 m/s
(E) 196 m/s
N = mg
μs N = mv2/r
v = sqrt(μsgr) = sqrt (0.4*9.8*50) = 14.0 m/s
10. A 90 kg firefighter needs to climb the stairs of a 20-m-tall building while carrying a 40
kg backpack filled with gear. How much power does he need to reach the top in 55 s?
a) 390 W b) 460 W c) 640 W d) 805 W e) 1024 W
=
P
W mg ∆y (90 kg + 40 kg)(9.8 m/s 2 )(20 m)
=
=
= 460 W
∆t
∆t
55 s
11. A particle moving along the x-axis experiences the force shown in the graph. If the
particle has 15 J of kinetic energy as it passes x = 4 m, what is its kinetic energy when it
reaches x = 10 m?
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a. 30 J
b. 24 J
c. 39 J
d. 45 J
K(x = 10 m) – K(x = 4 m) = (8-4)*6/2 + (-3)*(10-8)/2
K(x = 10 m) = K(x = 4 m) + 12 – 3 = 15+12 - 3 = 24 J
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e. 33 J
12. A human cannonball uses a spring-loaded cannon to launch himself. He has a mass of 70
kg, and the spring compresses a total distance of 1.40 m. If he needs a launch velocity of
18.0 m/s, what should be the value of the spring’s spring constant? The launch will result
in the human cannonball being launched horizontally.
a) 8.6 x 10-5 N/m
b) 11,600 N/m
c) 12,300 N/m
d) 14,672 N/m
e) 9.3 µN/m
13. A force F = {12i + 35j} N acts on a particle. This results in a displacement defined by Δr
= {20i + 15j} cm. What is the work done on this particle by this force?
a) 1.25 J
b) 6.25 J
c) 12.25 J
d) 7.65 J
e) 1,645 J
Solution
Δr = {20i + 15j} cm = {0.20i + 0.15j} m
W = F . Δr = (12i + 35j) . (0.20i + 0.15j) = (12i).(0.20i) + (35j).(0.15j) = 2.4 + 5.25 = 7.65 J
14. The figure to the right shows the potential
energy of a particle as a function of
position for a particle moving along the x
axis. Assume the corresponding force is
the only force acting on the particle.
Which of the following is true as the
particle moves from 𝑥𝑥𝐵𝐵 to 𝑥𝑥𝐶𝐶 ?
a. The force does positive work and the
kinetic energy decreases
b. The force does negative work and the
kinetic energy decreases
𝑥𝑥𝐴𝐴
𝑥𝑥𝐵𝐵
c. The force does positive work and the
Position, 𝑥𝑥
kinetic energy increases
d. The force does negative work and the kinetic energy increases
𝑥𝑥𝐶𝐶
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