Question

A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction = 0.1. Compute the
(a) work done by the applied force in 10 s.
(b) work done by friction in 10 s,
(c) work done by the net force on the body in 10 s,
(d) change in kinetic energy of the body in 10 s,
and interpret your results.

Answer 1

Given:

The mass of the body is $m=2kg$

The force applied on the body $F=7N$

The coefficient of kinetic friction $=0.1$

Since the body starts from rest, the initial velocity of body is zero.

Time at which the work is to be determined is $t=10s$

The acceleration produced in the body by the applied force is given by Newtons second law of motion as:

$a^{\prime }=\frac{F}{m}=\frac{7}{2}=3.5m/s^2$

Frictional force is given as:

$f=\mu g=0.1\times 2\times 9.8=1.96$

The acceleration produced by the frictional force:

$a"=-\frac{1.96}{2}=-0.98m/s^2$

Therefore, the total acceleration of the body:

$a^{\prime }+a"=3.5+(-0.98)=2.52m/s^2$

The distance traveled by the body is given by the equation of motion:

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

$=0+\frac{1}{2}\times 2.52\times (10{)}^2=126m$

(a) Work done by the applied force,

$W_a=F\cdot s=7\times 126=882J$

(b) Work done by the frictional force,

$W_f=F\cdot s=1.96\times 126=247J$

(c), (d)

From the first equation of motion, final velocity can be calculated as:

$v=u+at$

$=0+2.52\times 10=25.2m/s$

So, the change in kinetic energy is

$\Delta K=\frac{1}{2}m{v}^2-\frac{1}{2}m{u}^2$

$=\frac{1}{2}2(v^2-u^2)=(25.2{)}^2-0^2=635J$

The distance traveled by the body is given by the equation of motion:

$s=ut+\frac{1}{2}a{t}^2=0+\frac{1}{2}\times 2.52\times (10{)}^2=126m$

Net force $=7+\left(1.96\right)=5.04N$

Work done by the net force,

$W_{net}=5.04\times 126=635J$