Question

A body of mass 0.5 kg travels in a straight line with velocity v = ax512 where a = 5 in SI unit What is the work done by the net force during its displacement from x = 0 to x = 2 m?​

Answer 1

        $\text{Mass of body}(m) = 0.5\: kg$

        $\text{Acceleration}(a) = 5 \: ms^{-2}$

        $\text{Velocity}(v) = ax^{\dfrac32}$

$\implies \text{Velocity at }x = 0(v_0) = 5(0) = 0 \: ms^{-1}$

$\implies \text{Velocity at }x = 2 \: m (v_2) = 5(2)^{\dfrac{3}{2}} = 5\times2\sqrt2 = 10\sqrt2 \: ms^{-1}$

        $\text{We know that Work done} = \text{Change in K.E}$

$\implies \text{K.E}_i = \dfrac12m(v_0)^2 = 0 \: J$

$\implies \text{K.E}_f = \dfrac12m(v_2)^2 = \dfrac12\times 0.5 \times 200 = 50 \: J$

$\implies \text{Change in K.E}(\triangle{\text{K.E}}) = \text{K.E}_f - \text{K.E}_i$

$\implies \: \triangle{\text{K.E}} = 50 \: J$

$\implies \boxed{\text{W} = 50 \: J}$