Mass of the body, \(m=2 \mathrm{~kg}\)
Applied force, \(F=7 \mathrm{~N}\)
Coefficient of kinetic friction, \(\mu=0.1\)
Initial velocity, \(u=0\)
Time, \(t=10 \mathrm{~s}\)
The acceleration produced in the body by the applied force is given by Newton's
second law of motion as:
a' = F/m \(=7 / 2=3.5 \mathrm{~m} / \mathrm{s}^{2}\)
Frictional force is given as:
\(f=\mu \mathrm{mg}\)
\(=0.1 \times 2 \times 9.8=-1.96 \mathrm{~N}\)
The acceleration produced by
the frictional force:
\(a^{\prime}=-\frac{1.96}{2}=-0.98 \mathrm{~m} / \mathrm{s}^{2}\)
Total acceleration of the body:
\(a=a^{\prime}+a^{\prime}\)
\(=3.5+(-0.98)=2.52 \mathrm{~m} / \mathrm{s}^{2}\)
The distance travelled by the body is given
by the equation of motion:
$$
\begin{aligned}
&s=u t+\frac{1}{2} a t^{2} \\
&=0+\frac{1}{2} \times 2.52 \times(10)^{2}=126 \mathrm{~m}
\end{aligned}
$$
Work done by the applied force,
$$
W_{\mathrm{a}}=F \times s=7 \times 126=882 \mathrm{~J}
$$
Work done by the frictional force,
$$
W_{\mathrm{f}}=F \times s=-1.96 \times 126=-247 \mathrm{~J}
$$
Net force \(=7+(-1.96)=5.04 \mathrm{~N}\)
Work done by the net force, \(W_{\text {net }}=5.04 \times 126=635 \mathrm{~J}\)
From the first equation of motion, final velocity can be calculated as:
$$
\begin{aligned}
&v=u+a t \\
&=0+2.52 \times 10=25.2 \mathrm{~m} / \mathrm{s}
\end{aligned}
$$
$$
\begin{aligned}
&\text { Change in kinetic energy }=\frac{1}{2} m v^{2}-\frac{1}{2} m u^{2} \\
&=\frac{1}{2} \times 2\left(v^{2}-u^{2}\right)=(25.2)^{2}-0^{2}=635 \mathrm{~J}
\end{aligned}
$$
