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

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

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