Download W = mg W g = m = 1500 9.8 =153.06kg

SAMPLE EXAM 3 SOLUTIONS
PHYS 1114 Spring 2013
Prof. K. Tapp
1.
A student pulls a box of books of total mass 40kg on a smooth horizontal
floor with a 100-N force in a direction of 37° above the horizontal surface.
Find the components of the student’s pulling force.
2.
What is the mass of a 1500 N x-ray machine in a location on Earth?
W = mg
W
1500
=m=
= 153.06kg
g
9.8
3.
If the kinetic coefficient of friction between ice skates and ice is 0.01, what is the
average forward force necessary for a 500 N skater to maintain a constant
forward velocity?
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4.
A block is projected with a speed of 2.5 m/s on a horizontal surface. If the block
comes to rest in 1.5 m, what is the coefficient of kinetic friction between the block
and the surface?
5.
At a recent performance of “Peter Pan” in the OCCC Performing Arts Theater,
stage crews were able to make the lead actor fly on stage with the help of a
pulley system. Assuming a 250 N actor is lifted at a constant rate using a single
frictionless pulley, with how much force is the stage crews pulling on the cable?
We focus on Peter Pan and the forces that balance to keep
him in constant motion, which means a=0. Recall that
Tension is a constnat Force throughout the entire cable.
FNET = T − W
0 = T − 250N
T = F = 250N
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6.
Calculate the acceleration of the system if m1 = 6 kg, m2 = 10 kg and µk = 0.25
between m1 and the horizontal surface. The string is massless, and the pulley is
ideal and frictionless.
7.
An airplane has a mass of 3.1 x 104 kg and takes off under the influence of a
constant net force of 3.7 x 104 N. What is the net force that acts on the plane’s
78-kg pilot?
According to Newton’s second law, the acceleration is a = ΣF/m. Since the pilot and the plane have the
same acceleration, we can write
Therefore, we find
8.
In the amusement park ride known as Magic Mountain Superman, powerful
magnets accelerate a car and its riders from rest to 45 m/s (about 100 mi/h) in a
time of 7.0 s. The combined mass of the car and riders is 5.5 x 103 kg. Find the
average net force exerted on the car and riders by the magnets.
According to Newton’s second law, the average net force
is equal to the product of the object’s mass
m and the average acceleration . The average acceleration is equal to the change in velocity divided by
the elapsed time, where the change in velocity is the final velocity v minus the initial velocity v0.
The average net force exerted on the car and riders is
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9.
When a parachute opens, the air exerts a large drag force on it. This upward
force is initially greater than the weight of the sky diver and, thus, slows him
down. Suppose the weight of the sky diver is 915 N and the drag force has a
magnitude of 1027 N. The mass of the sky diver is 93.4 kg. What are the
magnitude and direction of his acceleration?
The acceleration of the sky diver can be obtained directly from Newton’s second law as the net force
divided by the sky diver’s mass. The net force is the vector sum of the sky diver’s weight and the drag
force. From Newton’s second law,
, the sky diver’s
acceleration is
f
f
The free-body diagram shows the two forces acting on the sky diver, his weight W and
the drag force f. The net force is
. Thus, the acceleration can be written
as
+
The acceleration of the sky diver is
−
W
Free-body Diagram
Note that the acceleration is positive, indicating that it points
10.
.
Two skaters, a man and a woman, are standing on ice. Neglect any friction
between the skate blades and the ice. The mass of the man is 82 kg, and the
mass of the woman is 48 kg. The woman pushes on the man with a force of 45 N
due east. Determine the acceleration (magnitude and direction) of (a) the man
and (b) the woman.
Since there is only one force acting on the man in the horizontal direction, it is the net force. According to
Newton’s second law, the man must accelerate under the action of this force. When the woman exerts a
force on the man, the man exerts a force of equal magnitude, but opposite direction, on the woman
(Newton’s third law). It is the only force acting on the woman in the horizontal direction, so, as is the case
with the man, she must accelerate.
The acceleration of the man is equal to the net force acting on him divided by his mass.
The acceleration of the woman is equal to the net force acting on her divided by her mass.
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11.
Determine the work done by a 50 N force which
displaces the block 6 ft.
12.
How much work is done in each case?
a)
A 40 kg block is pushed 36 m at constant velocity across a level surface
where the coefficient of kinetic friction is 0.22.
b)
A certain spring is extended from x = 0 to x = 4 cm. It was previously
found that a force of 70 N was required to extend the spring from x = 0 to x
= 2 cm.
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13.
A 0.25 kg baseball moves toward a batter with a velocity of 14 m/s. It is struck
with a bat which causes it to move in the opposite direction at 28 m/s. Find the
average force exerted on the ball if the bat and ball are in contact for 0.010 s.
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