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?

Answer 1

\[\text{ Given } : \]

\[\text{ Mass of the block, m = 250 gm = 0 . 250 kg } \]

\[\text{ Initial speed of the block, u = 40 cm/s = 0 . 4 m/s } \] 

\[\text{ Final speed of the block, }\nu = 0 \]

\[\text{ Coefficient of friction, }  \mu = 0 . 1\]

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,

\[W = - \mu \text{ Rs }  \cos \theta\]

\[ = - 1 \times 2 . 5 \times 0 . 082 \times 1\]

\[ = - 0 . 02 J\]

\[ \Rightarrow W = - 0 . 02 J\]