Question 214: Work, Energy and Power & Circular Motion

Question

If a body looses half of its velocity on penetrating $$3$$ $$cm$$ in a wooden block, then how much will it penetrate more before coming to rest?

Accepted Answer

We know the work energy theorem, $$W = \Delta K = FS$$
For first penetration, by applying work energy theorem we get,
$${1 \over 2}m{v^2} - {1 \over 2}m{\left( {{v \over 2}} ight)^2} = F 	imes 3\,\,...(i)$$
For second penetration, by applying work energy theorem we get,
$${1 \over 2}m{\left( {{v \over 2}} ight)^2} - 0 = F 	imes S\,...(ii)$$
On dividing $$(ii)$$ by $$(i)$$
$${{1/4} \over {3/4}} = S/3$$
$$	herefore$$ $$S = 1\,cm$$

Answer

$$1$$ $$cm$$

Answer

$$2$$ $$cm$$

Answer

$$3$$ $$cm$$

Answer

$$4$$ $$cm$$

Work, Energy and Power & Circular Motion

If a body looses half of its velocity on penetrating $3 cm$ in a wooden block, then how much will it penetrate more before coming to rest?

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