Explanation
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 mass of the bucket $$m=20\ kg$$
The height of the building $$h=20\ m$$
$$\text{Work done by the worker} = \text{Potential energy of bucket at the top of the building}$$
$$\therefore W = mgh$$
$$\Rightarrow W = 20 \times 9.8 \times 20$$
$$\Rightarrow W = 3920\;{\rm{J}}$$
Thus, the required work done is $$3.92\;k{\rm{J}}$$
K.E$$_1$$ = K.E$$_2$$
$$\dfrac{1}{2}$$ $$M_1$$ $$v_1^2$$ = $$\dfrac{1}{2}$$ $$M_2$$ $$v_2^2$$
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