Question

A $50\text{kg}$ block slides down an incline with an angle of incline ${45}^{\circ }$. The coefficient of friction between the block and the incline is $0.3$. A worker applies a force on the block, so that it slides down the incline of length $2\text{m}$ with a constant velocity. What is the work done by the worker on the block?

• A. $495.2\text{J}$
• B. $-495.2\text{J}$
• C. $-525.7\text{J}$
• D. $525.7\text{J}$

Answer 1

The correct option is B $-495.2\text{J}$


From the FBD,
$N=mg\cos \theta$
and frictional force,
$F_f=\mu mg\cos \theta =0.3\times 50\times 10\times \cos {45}^{\circ }$
$\Rightarrow F_f=106\text{N}$
The force that causes the block to slide down the incline
$F_i=mg\sin \theta =50\times 10\times \sin {45}^{\circ }$
$\Rightarrow F_i=353.6\text{N}$
Net force on the block,
$F_i-F_f=353.6-106=247.6\text{N}$

The worker counters this force so that there is no net acceleration.
$\therefore$ The work done by the worker on the block
$W=-F\times d$
$\Rightarrow W=-247.6\times 2=-495.2\text{J}$
(sign is negative because force applied by worker is opposite to displacement of block)