# A cart of mass M1 = 6 kg is attached to a block of mass M2 = 3 kg by a string that passes over a frictionless pulley. The system is initially at rest and the table is frictionless. After the block has fallen a distance h = 1 m:

What is the work Ws done on the cart by the string?

(the cart is on the table attached to a pulley from which a block is hanging)

After the block has fallen 1 m, potential energy of M2 g H = 3*9.8*1 = 29.4 J is converted to kinetic energy of both the cart and the block. Since they will both be travelling at the same speed, and the cart has 2/3 of the total mass, only 2/3 of the potential energy loss acts to accelerate the cart. That would be 19.6 J.

You could get the same result by solving two free body equations to get the tension in the string, and multiplying it by the distance moved.

T = M1 g = M2 a

M2 g -T = M1 a

M2 g = (M1 + M2) a

a = g [M2/(M1 + M2)]

T = M2 a = M2 g * [M2/(M1 + M2)]

= 2/3 * M2 g

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