Question

A $60.0-kg$ person running at an initial speed of $4.00m/s$ jumps onto a $120-kg$ 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.

Answer 1

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

$60\times 4=(120+60)vv=\frac{60\times 4}{120+60}v=\frac{240}{180}v=\frac{4}{3}m/s$

Change in kinetic energy of cart is

$△KE=\frac{1}{2}M{v}^2-\frac{1}{2}M{u}^2$

Substitute all value in above equation

$△KE=\frac{1}{2}M(v^2-u^2)△KE=\frac{1}{2}\times 120\times ({\frac{4}{3}}^2-0)△KE=60\times \frac{16}{9}△KE=106.67J$