Explanation
It is given that 2 kg and 3 kg blocks are placed on a smooth horizontal surface and connected by spring which is unstretched initially. Let v is the final velocity then from conservation of momentum:
$$ 5v=2(-4{{v}_{0}})+3({{v}_{0}}) $$
$$ 5v=-8{{v}_{0}}+3{{v}_{0}} $$
$$ v=-{{v}_{0}}...........(1) $$
Let x is the maximum elongation in the spring. From conservation of energy:
$$ \dfrac{1}{2}(2){{(-4{{v}_{0}})}^{2}}+\dfrac{1}{2}3{{({{v}_{0}})}^{2}}=\dfrac{1}{2}k{{x}^{2}}+\dfrac{1}{2}5{{({{v}_{0}})}^{2}}\,(from\,\,equation(1)) $$
$$ 16{{v}_{0}}^{2}+\dfrac{3}{2}v_{0}^{2}=\dfrac{1}{2}k{{x}^{2}}+\dfrac{5}{2}v_{0}^{2} $$
$$ 15v_{0}^{2}=\dfrac{1}{2}k{{x}^{2}} $$
The energy stored in spring is $$15v_{0}^{2}$$
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$$_2$$
$$\dfrac{1}{2}$$ $$M_1$$ $$v_1^2$$ = $$\dfrac{1}{2}$$ $$M_2$$ $$v_2^2$$
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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