Question

A block of mass $m$ is moving down with constant velocity along an inclined plane of inclination $\theta$. What is the work done by an external force (applied parallel to the inclined plane) in pulling the block up the inclined plane through a height ${}^{\prime }{h}^{\prime }$ with constant velocity?

• A. $3mgh$
• B. $mgh$
• C. $mgh\sin \theta$
• D. $2mgh$

Answer 1

The correct option is D $2mgh$


Since the block moves down along the inclined plane with constant velocity, its net acceleration is zero. Therefore. the net external force acting on the block is zero.
$\Rightarrow \sum F_x=0;\sum F_y=0\Rightarrow mg\sin \theta -f_k=0$ $\Rightarrow f_k=mg\sin \theta \dots \dots \left(1\right)$

Let the applied force $F$ be parallel to the inclined plane. When it pulls the block up, kinetic friction acts down to oppose the relative motion.

The work done by the force to displace the block through a distance $l$ along the inclined plane is given as
$W=F.s=Fl$ where $l=hcosec\theta \Rightarrow W=Fhcosec\theta \dots \dots \left(2\right)$
Since the block slides up with constant velocity, therefore the net force acting on the block
$\sum F_x=0;\sum F_y=0\Rightarrow mg\sin \theta +f_k=F\dots \dots \left(3\right)$
Using $\left(1\right)$ and $\left(3\right)$
$F=2mg\sin \theta \dots \dots \left(4\right)$
Using $2$ and $3$ we get
$W=2mgh$