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.
Determine the displacement of the person relative to the ground while he is sliding on 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

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

We know that change in momentum is impulse that is force multiplied by time .The force which applied to increase the motion.

For person,

$mv-mu=Im(u-v)=F_k\times t$ ..............................(1)

We know that friction force is

$F_k=\mu N$

In any object weight $\left(mg\right)$ always balance the normal force therefore we can replace normal force by weight.

$F_k=\mu mg$

Substitute all value in above equation

$F_k=0.4\times 60\times 9.8F_k=235.2N$

Substitute all value in equation (1)

$70(4-\frac{4}{3})=235.2\times t70\times \frac{12-4}{3}=235.2\times t\frac{70\times 8}{3}=235.2\times tt=\frac{70\times 8}{3\times 235.2}t=\frac{560}{705.6}t=0.79s$

displacement of the person relative to the ground while he is sliding on the cart is

$\frac{1}{2}(v+u)t=\frac{1}{2}(\frac{4}{3}+4)0.79=2.11m$