Explanation
For $$(A):$$ Displacement $$< 0$$ and $$F<0$$ so $$W>0$$
For $$(B):$$ Displacement $$>0$$ and $$F>0$$ so $$W>0$$
For $$(C):$$ Displacement $$>0$$ and $$F<0$$ so $$W<0$$
For $$(D):$$ $$W=\int _{ -2 }^{ 2 }{ -6{ x }^{ 3 } } dx=\left( -\cfrac { 6 }{ 4 } \right) { \left( { x }^{ 4 } \right) }_{ -2 }^{ 2 }=0$$
Given:
$$m_{A} = 20 kg$$
$$m_{B} = 5 kg$$
Kinetic Energy of both bodies A and B are the same.
Thus,
$$\dfrac{1}{2} m_{A} V_{A}^{2} = \dfrac{1}{2} m_{B} V_{B}^{2} \\$$
$$\dfrac{V_{A}^{2}}{V_{B}^{2}} = \dfrac{m_{B}}{m_{A}} \\$$
$$\dfrac{V_{A}^{2}}{V_{B}^{2}} = \dfrac{5}{20}\\$$
$$\dfrac{V_{A}}{V_{B}} = \dfrac{1}{2}$$
So, option A is correct.
The amount of the work done in raising the glass of the water through a height is calculated by the amount of potential energy
$$W = mgh$$
By substituting the value in the above equation we get
$$W = 0.5 \times 10 \times 0.5$$
Hence the work done in raising the glass is$${\rm{2}}{\rm{.5J}}$$
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