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Standard XII

Physics

Block on a Block Problems

The block of ...

Question

The block of mass $1 k g$ slides down a curved track which forms one quadrant of a circle of radius $1 m$ as shown in the figure. The speed of block at the bottom os the track is $v = 2$ $m s^{- 1}$ . The work done by the force of friction is:

+ 4 J

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- 4 J

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- 8 J

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+ 8 J

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Solution

The correct option is C $- 8 J$
Work done by the friction is the change in energy of the block.
Work done by friction $= K . E - P . E = \frac{1}{2} m v^{2} - m g h$
$\Rightarrow$ work done $= (0.5) (1) (2^{2}) - (1) (10) (1) = - 8 J$

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The block of mass 1 kg slides down a curved track which forms one quadrant of a circle of radius 1 m as shown in the figure. The speed of block at the bottom os the track is v=2 ms−1. The work done by the force of friction is:

The correct option is C −8 JWork done by the friction is the change in energy of the block.Work done by friction=K.E−P.E=12mv2−mgh⇒ work done=(0.5)(1)(22)−(1)(10)(1)=−8 J

The block of mass $1 k g$ slides down a curved track which forms one quadrant of a circle of radius $1 m$ as shown in the figure. The speed of block at the bottom os the track is $v = 2$ $m s^{- 1}$ . The work done by the force of friction is:

The correct option is C $- 8 J$
Work done by the friction is the change in energy of the block.
Work done by friction $= K . E - P . E = \frac{1}{2} m v^{2} - m g h$
$\Rightarrow$ work done $= (0.5) (1) (2^{2}) - (1) (10) (1) = - 8 J$