# A body of mass 2 kg initially at rest moves under the action

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,

solution:

Mass of the body, m = 2 kg

Applied force, F = 7 N

Coefficient of kinetic friction, $\mu=0.1$

Initial velocity, $u=0$

$\operatorname{Time} t=10 c$

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

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

Frictional force is given as:

$f=\mu m g$

$=0.1 \times 2 \times 9.8=-1.96 \mathrm{~N}$

The acceleration produced by the frictional force:

$a^{\prime \prime}=-\frac{1.96}{2}=-0.98 \mathrm{~m} / \mathrm{s}^{2}$

Total acceleration of the body:

$a=a^{\prime}+a^{\prime \prime}$

$=3.5+(-0.98)=2.52 \mathrm{~m} / \mathrm{s}^{2}$

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

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

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

(a) Work done by the applied force. $W_{3}=F \times s=7 \times 126=882 \mathrm{~J}$

(b) Work done by the frictional force, $W_{\mathrm{f}}=F \times s=-1.96 \times 126=-247 \mathrm{~J}$

(c) Net force $=7+(-1.96)=5.04 \mathrm{~N}$

Work done by the net force, $W_{\text {net }}=5.04 \times 126=635 \mathrm{~J}$

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

v = u + at

$=0+2.52 \times 10=25.2 \mathrm{~m} / \mathrm{s}$

Change in kinetic energy $=\frac{1}{2} m v^{2}-\frac{1}{2} m u^{2}$

$=\frac{1}{2} \times 2\left(v^{2}-u^{2}\right)=(25.2)^{2}-0^{2}=635 \mathrm{~J}$