Question

A block of mass 1.970 kg is free to slide on a frictionless, horizontal surface. A cord attached to the block passes over a pulley whose diameter is 0.150 m , to a hanging book with mass 3.030 kg . The system is released from rest, and both the book and the block are observed to move a distance 1.19 m over a time interval of 0.780 s ; at the instant at which that distance is measured, the book is still moving downwards and has not yet contacted the floor beneath it. The pulley is not massless, but it is frictionless.

During the action described in the introduction, what is the work done on the pulley by the force on the pulley by the horizontal part of the cord? Also, what is the work done on the pulley by the force on the pulley by the vertical part of the cord?

Answer #1

consider the motion of the book

v_{o} = initial velocity = 0 m/s

d = distance traveled = 1.19 m

t = time taken = 0.780 s

a = acceleration = ?

Using the equation

d = v_{o} t + (0.5) a t^{2}

1.19 = (0) (0.780) + (0.5) a (0.780)^{2}

a = 3.91 m/s^{2}

T_{1} = tension in horizontal part of the cord

T_{2} = tension in vertical part of the cord

For the block, force equation is given as

T_{1} = m a

T_{1} = (1.970) (3.91) = 7.7 N

For the book, force equation is given as

Mg - T_{2} = Ma

(3.03 x 9.8) - T_{2} = (3.03) (3.91)

T_{2} = 17.85 N

Work done by horizontal part is given as

W_{1} = - T_{1} d = - 7.7 x 1.19 = 9.2 J

Work done by vertical part is given as

W_{2} = T_{2} d = 17.85 x 1.19 = 21.2 J

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