Given, $m = 2 \text{ kg}, u = 0$, $m = 0.1, t = 10 \text{ s}$ Applied force $F = 7 \text{ N}$ Force due to friction $f = \mu m g = 0.1 \times 2 \times 9.8 = 1.96 \text{ N}$ Net force under which body moves $F' = F - f = 7 - 1.96 = 5.04 \text{ N}$ Therefore acceleration with which body moves $a = \frac{F'}{m} = \frac{5.04}{2} = 2.52 \text{ m s}^{-2}$Therefore distance moved by the body in 10 seconds$S = u t + \frac{1}{2} a t^{2}$ $= 0 \times 10 + \frac{1}{2} \times 2.52 \times 10^{2}$ $= 126 \text{ m}$(a) Work done by the applied force in 10 s$W = F \times S = 7 \times 126 = 882 \text{ J}$(b) Work done by the friction force in 10 s$W = - f \times S = -1.96 \times 126$ $= -246.9 \text{ J} = -247 \text{ J}$(c) Work done by the net force on body in 10 s$W = F' \times S = 5.04 \times 126 = 635 \text{ J}$(d) Initial kinetic energy$= \frac{1}{2} m u^{2} = \frac{1}{2} m \times (0)^{2} = 0$Final kinetic energy$= \frac{1}{2} m v^{2} = \frac{1}{2} \times 2 \times (25.2)^{2} = 635 \text{ J}$Change in kinetic energy = Final K.E. – Initial K.E. = 635 – 0 = 635This shows that change in kinetic energy of the body is equal to work done by net force on the body.