A bowling ball of mass m=5 kg hangs from a massless string of length L=10.0 m fr
ID: 2132040 • Letter: A
Question
A bowling ball of mass m=5 kg hangs from a massless string of
length L=10.0 m from the ceiling of a large lecture hall. With the ball
starting from rest and hanging straight down, the professor pushes
horizontally on the ball with a varying force F, to move the bowling
ball a distance d=3.50 m to the side.
(a) What is the magnitude of the horizontal force supplied by the
professor on the bowling ball at the end of the displacement where
the ball is again motionless?
(b) During the time that the ball is being displaced, what is the total
work done on the bowling ball?
(c) During the time that the ball is being displaced, what is the work
done by the gravitational force?
(d) During the time that the ball is being displaced, what is the work done on the ball by the tension
in the rope?
(e) Knowing that the ball is motionless before and after its displacement, use the answers to (b), (c),
and (d) to find the work the professor
A bowling ball of mass m=5 kg hangs from a massless string of length L=10.0 m from the ceiling of a large lecture hall. With the ball starting from rest and hanging straight down, the professor pushes horizontally on the ball with a varying force F, to move the bowling ball a distance d=3.50 m to the side. What is the magnitude of the horizontal force supplied by the professor on the bowling ball at the end of the displacement where the ball is again motionless? During the time that the ball is being displaced, what is the total work done on the bowling ball? During the time that the ball is being displaced, what is the work done by the gravitational force? During the time that the ball is being displaced, what is the work done on the ball by the tension in the rope? Knowing that the ball is motionless before and after its displacement, use the answers to (b), (c), and (d) to find the work the professor's force, F, does on the bowling ball. Why is the work of the professor's force not equal to the product of the horizontal displacement and the answer to part (a)?Explanation / Answer
tan theta = 3.5 / 10 = .35 theta = 19.3 deg
T cos theta = m g vertical component of tension
T sin theta = F horizontal component of tension
F = m g tan theta dividing equations
(a) F = 5 * 9.8 * .35 = 17.2 N
(b) W = m g h = m g (L - L cos theta) distance ball is raised
W = 5 * 9.8 * 10 * (1 - cos 19.3) = 27.5 J
(c) -27.5 J = Work done by gravity (due to elevation of ball)
(d) The tension does no work because it is allways perpendicular
to the displacement of the ball
(e) The work done by the professor is 27.5 J
(f) F = m g tan theta so the force the professor applies varies
from zero to the final displacement
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