A body of mass 2 kg is dropped from a building of height 5 m from the ground. The maximum kinetic energy the body can attain is: (Assume $g = 10 m / s^{2}$ )

50 J

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100 J

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150 J

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200 J

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Solution

The correct option is B
100 J

From $1^{s t}$ equation of motion we have, $v^{2} = u^{2} + 2 g h$

As it is dropped so, $u = 0$ and $g = 10 m / s^{2}$ and $h = 5 m$

Putting these values we get, $v = 10 m / s$

So, the body would attain the maximum kinetic energy when it strikes the ground, i.e. K.E = $\frac{1}{2} \times m \times v^{2} = \frac{1}{2} \times 2 \times 10^{2} = 100 J$

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A body of mass 2 kg is dropped from a building of height 5 m from the ground. The maximum kinetic energy the body can attain is: (Assume g=10m/s2)

The correct option is B 100 J From 1st equation of motion we have, v2=u2+2gh As it is dropped so, u=0 and g=10 m/s2 and h=5 m Putting these values we get, v=10 m/s​ So, the body would attain the maximum kinetic energy when it strikes the ground, i.e. K.E = 12×m×v2=12×2×102=100 J

A body of mass 2 kg is dropped from a building of height 5 m from the ground. The maximum kinetic energy the body can attain is: (Assume $g = 10 m / s^{2}$ )