Explanation
Given:
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$$_2$$
$$\dfrac{1}{2}$$ $$M_1$$ $$v_1^2$$ = $$\dfrac{1}{2}$$ $$M_2$$ $$v_2^2$$
Given that,
Spring constant $$k=100$$
Mass $$m=10\,kg$$
Stretching $$x=2\,m$$
We know that,
The work done is
$$ W=\dfrac{1}{2}k{{x}^{2}} $$
$$ W=\dfrac{1}{2}\times 100\times 4 $$
$$ W=200\,J $$
Hence, the work done is $$200\ J$$
Two identical balls are released from positions as shown. They collide elastically on horizontal surface. Ratio of heights attained by A & B after collision is( All surface are smooth,neglect energy loss at M & N)
Work done in stretching wire
$$=\dfrac {1}{2}\dfrac {YAL^{2}}{L}=\dfrac {1}{2}F.l$$
Where
$$L=$$ length of wire
$$l=$$ increase in length
We know that
$$E=\dfrac {1}{2}\times F \times AL$$
$$=\dfrac {1}{2}\times 200 \times 10^{-3}$$
$$=0.1J$$
Hence $$(A)$$ option is correct.
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