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Standard XII

Physics

Work Done as Dot Product

A body of mas...

Question

A body of mass 0.1 kg is dropped from a height of 10 m at a place wheres $g = 10 m s^{- 2}$ . Its KE just before its strikes the ground is:

1 J

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1.04 J

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3.5 J

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10 J

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Solution

The correct option is D 10 J
Here, by conservation of energy, the potential energy of the body at height $h$ is converted into kinetic energy before its strikes the ground.
Thus, $m g h = K E$
or $K E = 0.1 (10) 10 = 10 J$

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A body of mass 0.1 kg is dropped from a height of 10 m at a place wheres g=10ms−2. Its KE just before its strikes the ground is:

The correct option is D 10 JHere, by conservation of energy, the potential energy of the body at height h is converted into kinetic energy before its strikes the ground. Thus, mgh=KEor KE=0.1(10)10=10J

A body of mass 0.1 kg is dropped from a height of 10 m at a place wheres $g = 10 m s^{- 2}$ . Its KE just before its strikes the ground is:

The correct option is D 10 J
Here, by conservation of energy, the potential energy of the body at height $h$ is converted into kinetic energy before its strikes the ground.
Thus, $m g h = K E$
or $K E = 0.1 (10) 10 = 10 J$