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. $\mu mgh$
• B. $\mu mgh/\sin \theta$
• C. $mgh$
• D. $\mu mgh/\cos \theta$

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 $W_{\text{friction }}=-mgh$
So, energy dissipated due to friction $=mgh$.