Question

A block of mass $m$ is sliding down an inclined plane with constant speed.At a certain instant $t_{0}$, its height above the ground is $h$. The coefficient of kinetic friction between the block and the plane is $\mu$. If the block reaches the ground at a later instant $t_{g}$, then the energy dissipated by friction in the time interval $\left(t_{g}-t_{0}\right)$ is

• A. $\mu m g h$
• B. $\mu m g h / \sin \theta$
• C. $m g h$
• D. $\mu m g h / \cos \theta$

Answer 1

Answer: (C)

As block is sliding with a constant speed,
so change in kinetic energy of block when it reaches bottom is zero.
Now, by work-energy theorem,
Total work done $=$ Change in kinetic energy
$\Rightarrow W_{\text {friction }}+W_{\text {gravitation }}=\Delta KE$
$\Rightarrow W_{\text {friction }}=-W_{\text {gravitation }}$
or $ W_{\text {friction }}=-m g h$
So, energy dissipated due to friction $=m g h$.