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Standard XII

Physics

Work Energy Theorem Application

A block of ma...

Question

A block of mass $1 k g$ slides down a curved track that is one quadrant of a circle of radius $1 m$ . Speed of the block at the bottom is $2 m / s$ . Work done by the frictional force on the block when it reaches at the bottom is:

8 J

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- 8 J

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4 J

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0 J

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Solution

The correct option is A $- 8 J$
Given, $m = 1 k g$ ,
$v = 2 m / s$
$R = 1 m$
$W = ?$
From energy conservation,
$m g h + W = \frac{1}{2} m v^{2}$
$W = \frac{1}{2} m v^{2} - m g h$
$= \frac{1}{2} \times 1 (2)^{2} - 1 \times 10 = - 8 J$ .

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A block of mass 1 kg slides down a curved track that is one quadrant of a circle of radius 1 m. Speed of the block at the bottom is 2 m/s. Work done by the frictional force on the block when it reaches at the bottom is:

The correct option is A −8 JGiven, m=1 kg,v=2m/sR=1mW=?From energy conservation,mgh+W=12mv2W=12mv2−mgh=12×1(2)2−1×10=−8 J.

A block of mass $1 k g$ slides down a curved track that is one quadrant of a circle of radius $1 m$ . Speed of the block at the bottom is $2 m / s$ . Work done by the frictional force on the block when it reaches at the bottom is:

The correct option is A $- 8 J$
Given, $m = 1 k g$ ,
$v = 2 m / s$
$R = 1 m$
$W = ?$
From energy conservation,
$m g h + W = \frac{1}{2} m v^{2}$
$W = \frac{1}{2} m v^{2} - m g h$
$= \frac{1}{2} \times 1 (2)^{2} - 1 \times 10 = - 8 J$ .