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Question

A 60.0kg person running at an initial speed of 4.00 m/s jumps onto a 120kg cart initially at rest (Fig. P9.69). The person slides on the cart’s top surface and finally comes to rest relative to the cart. The coefficient of kinetic friction between the person and the cart is 0.400. Friction between the cart and ground can be ignored.
Find the change in kinetic energy of the cart.

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Solution

Given,
mass of person is 60kg
initial velocity is 4m/s
mass of cart is 120kg
Co-efficient of kinetic friction is 0.4
Let distance travel by the cart is x
final velocity when man reached the cart.
mu=(M+m)v
Substitute all value in above equation
60×4=(120+60)vv=60×4120+60v=240180v=43m/s
Change in kinetic energy of cart is
KE=12Mv212Mu2
Substitute all value in above equation
KE=12M(v2u2)KE=12×120×(4320)KE=60×169KE=106.67J

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