$A$ 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.

(D) Mass of the body,$m = 2 \; kg$
Applied force,$F = 7 \; N$
Coefficient of kinetic friction,$\mu = 0.1$
Initial velocity,$u = 0$
Time,$t = 10 \; s$
Acceleration due to applied force: $a_F = F/m = 7/2 = 3.5 \; m/s^2$
Frictional force: $f = \mu m g = 0.1 \times 2 \times 9.8 = 1.96 \; N$
Acceleration due to friction: $a_f = -f/m = -1.96/2 = -0.98 \; m/s^2$
Net acceleration: $a = a_F + a_f = 3.5 - 0.98 = 2.52 \; m/s^2$
Distance travelled: $s = ut + 0.5 a t^2 = 0 + 0.5 \times 2.52 \times 100 = 126 \; m$
$(a)$ Work done by applied force: $W_a = F \times s = 7 \times 126 = 882 \; J$
$(b)$ Work done by friction: $W_f = -f \times s = -1.96 \times 126 = -246.96 \; J \approx -247 \; J$
$(c)$ Work done by net force: $W_{net} = (F - f) \times s = 5.04 \times 126 = 635.04 \; J \approx 635 \; J$
$(d)$ Change in kinetic energy: $\Delta K = W_{net} = 635 \; J$
Interpretation: The work-energy theorem states that the net work done on a body equals the change in its kinetic energy.