Explanation
From work energy theorem,
work done by weight + work done by air$$=ΔKE$$ (change in Kinetic Energy)
$${ W }_{ air }=ΔKE−{ W }_{ (weight) }$$
$$=\dfrac{1}{2}m{ v }^{ 2 }−mgh$$
$$=\dfrac{1}{2}×5×{ (10) }^{ 2 }−5×10×20$$
$$=−750J$$
$$W = 200 \times 20 \times \dfrac{\sqrt3}{2}$$ $$\left( {\because \theta = 30^\circ ,\cos 30^\circ = \dfrac{\sqrt3}{2}} \right)$$
$$W = 3464J$$
Lets initial velocity of car is $$v$$, final velocity of this car will be $$5\ v$$
$$K.E = \dfrac{1}{2}$$ $$ mv ^{2}$$
$$\therefore \ K.E\propto v^2$$
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