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Question
A 250 g block slides on a rough horizontal table. Find the work done by the frictional force in bringing the block to rest if it is initially moving at a speed of 40 cm/s. If the friction coefficient between the table and the block is 0⋅1, how far does the block move before coming to rest?
Solution
\[\text{ Given } : \]
\[\text{ Mass of the block, m = 250 gm = 0 . 250 kg } \]
\[\text{ Initial speed of the block, u = 40 cm/s = 0 . 4 m/s } \]
\[\text{ Final speed of the block, }\nu = 0 \]
\[\text{ Coefficient of friction, } \mu = 0 . 1\]
Force in the forward direction is equal to the friction force.
\[\text{ Here } , \mu \text{ R = ma } \]
\[\left( \text{ where a is deceleration } \right)\]
\[a = \left( \frac{\mu R}{m} \right) = \left( \frac{\mu \text{ mg }}{m} \right) = \mu \text{ g } \]
\[ = 0 . 1 \times 9 . 8 = 0 . 98 \text{ m/ s} ^2 \]
\[s = \frac{\nu^2 - u^2}{2a} = 0 . 082 \text{ m } \]
\[ = 8 . 2 \text{ cm } \]
Again, work done against friction,
\[W = - \mu \text{ Rs } \cos \theta\]
\[ = - 1 \times 2 . 5 \times 0 . 082 \times 1\]
\[ = - 0 . 02 J\]
\[ \Rightarrow W = - 0 . 02 J\]
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