Given : $m=2 \mathrm{~kg}, u=0, F=7 \mathrm{~N}, \mu=0.1, t=10 \mathrm{~s}, W=$ ?
Acceleration produced by applied force
$$
=a_1=\frac{F}{m}=\frac{7}{2}=3.5 \mathrm{~m} / \mathrm{s}^2 \text {. }
$$
Force of friction $=\mathrm{f}=\mu \mathrm{R}=\mu \mathrm{mg}$
$$
=0.1 \times 2 \times 9.8=1.96 \mathrm{~N}
$$
Retardation produced by friction
$$
=a_2=\frac{-f}{m}=\frac{-1.96}{2}=-0.98 \mathrm{~m} / \mathrm{s}^2
$$
Net acceleration $=a=a_1+a_2=3.50-0.98$
$$
=2.52 \mathrm{~m} / \mathrm{s}^2
$$
Distance moved by the body in 10 seconds $\mathrm{S}=u t+\frac{1}{2} a t^2=0+\frac{1}{2} \times 2.52 \times(10)^2=126 \mathrm{~m}$.
(a) Work done by the applied force $=\mathrm{W}=\mathrm{FS}=7 \times$ $126=882 \mathrm{~J}$
(b) Work done by the frictional force $=-f \times s$ $=-1.96 \times 126=-246.9 \mathrm{~J}$
(c) Work done by the net force $=$ Net force $\times$ distance $=(7-1.96) \times 126=635 \mathrm{~J}$
(d) Velocity at the end of $10 \mathrm{~s}$
is $v=u+a t=0+2.52 \times 10=25.2 \mathrm{~m} / \mathrm{s}$
Final K.E. $=\frac{1}{2} m v^2=\frac{1}{2} \times 2 \times(25.2)^2=635 \mathrm{~J}$
Initial K.E. $=0$
$\therefore$ Change in $\mathrm{KE}=635-0=635 \mathrm{~J}$
This shows that the change in K.E. is equal to the work done by the net force.