Question

A block of mass \( M \) is pushed momentarily on a horizontal surface with initial velocity \( V \). If \( \mu \) is the coefficient of sliding friction between the block and surface, the block will come to rest after time \( t \) (where \( g \) is the acceleration due to gravity)

• A. \( \frac{\mu g}{V} \)
• B. \( \mu g V \)
• C. \( \frac{v^2}{\mu g} \)
• D. \( \frac{V}{\mu g} \)

Hint:

The time to come to rest for an object with initial velocity on a surface with friction is found by dividing the initial velocity by the frictional deceleration.

Answer 1

The Correct Option is D

Step 1: Understanding the motion of the block.
The block experiences a frictional force \( F_{\text{friction}} = \mu Mg \) that opposes its motion. This force decelerates the block, and we can use the equation of motion to find the time taken for the block to come to rest. Step 2: Using the equation of motion.
Using the equation \( v = u + at \), where: - \( u \) is the initial velocity \( V \), - \( v \) is the final velocity (0, since the block comes to rest), - \( a \) is the acceleration (deceleration in this case), - \( t \) is the time to come to rest. The frictional force provides a constant deceleration \( a = \frac{F_{\text{friction}}}{M} = \mu g \). The equation becomes: \[ 0 = V - \mu g t \quad \Rightarrow \quad t = \frac{V}{\mu g} \] Step 3: Conclusion.
Thus, the time taken for the block to come to rest is \( \frac{V}{\mu g} \), which corresponds to option (D).