Question

A rigid body of mass $2$kg initially at rest moves under the action of an applied horizontal force at $7$N on a table with coefficient of kinetic friction $=0.1$, Calculate the:

1. Work done by the applied force on the body in $10$s

2. Work done by friction on the body in $10$s

3. Work done by the net force on the body in $10$s

4. Change in kinetic energy of the body in $10$s

Answer 1

Given Mass $m=2kg,$ Force $F=7N$, Coefficient of kinetic friction $\mu =0.1$ ,initial velocity $u=0m/s$

According to Newton's second law:

$F=ma$

$F_{applied}-f=ma$ where f is force of friction.

$F_{applied}-{\mu }_{mg}=ma$

$7-0.1\times 2\times 10=2a$

$a=\frac{5}{2}=3m/s^2$

$s=ut+\frac{1}{2}a{t}^2$

$=0\times 10+\frac{1}{2}\times 3\times {10}^2=150m$

Work done by applied force:

$Wapplied=F\times s=7\times 150=1050J$

Work done by friction:

$W_{friction}=f\times s$

${\mu }_{mg}\times s$

$=0.1\times 2\times 10\times 150=300J$

Net work $W_{applied}-W_{friction}$

$=1050-300=750J$

This is also equal to change in kinetic energy due to work work energy theorem.