A Bowling ball of mass m=5kg hangs from a massless string of length L=10.0m from

Question

A Bowling ball of mass m=5kg hangs from a massless string of length L=10.0m from the cieling of a large lecture hall.  With the ball starting from rest and hanging 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's force, F, does on the bowling ball.

F) 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

it will be at angle given by sin theta = 3.5/10

theta =arcsin(3.5/10)=20.49 degrees

A) sum forces in the y

T cos theta - mg = 0

T = m g/cos theta

in the x

F - T sin theta = 0

F = m g tan(theta) = 5*9.81*tan(20.49)=18.33 N

B)W total = dKE = 0

C) W = m g dh = -5*9.81* (10-10*cos(20.49))=-524.6 J

D) W = F d cos phi, where phi is angle between F and d

here T is at 90 from movement so

W = 0

E) W professor = 524.6 J

F) since the movement is not paralell to the force.

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