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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.
Determine the displacement of the person relative to the ground while he is sliding on 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
We know that change in momentum is impulse that is force multiplied by time .The force which applied to increase the motion.
For person,
mvmu=Im(uv)=Fk×t ..............................(1)
We know that friction force is
Fk=μN
In any object weight (mg) always balance the normal force therefore we can replace normal force by weight.
Fk=μmg
Substitute all value in above equation
Fk=0.4×60×9.8Fk=235.2N
Substitute all value in equation (1)
70(443)=235.2×t70×1243=235.2×t70×83=235.2×tt=70×83×235.2t=560705.6t=0.79s
displacement of the person relative to the ground while he is sliding on the cart is
12(v+u)t=12(43+4)0.79=2.11m

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