Question

A block of mass 3 kg is moving down with constant velocity along a rough inclined plane. The work to be done by an external force in pulling the block along the inclined plane through a height of 50 cm is (Acceleration due to gravity $=10\,ms^{-2}$)

• A. 10 J
• B. 20 J
• C. 30 J
• D. 15 J

Hint:

Whenever a body moves with constant velocity, acceleration is zero and the net force acting on it must also be zero.

Answer 1

The Correct Option is D

Concept: When a body moves down an inclined plane with constant velocity, the net force acting along the plane is zero. Therefore, the frictional force balances the component of gravitational force along the plane. If the block is pulled upward through the same path at constant velocity, the external force must overcome both gravity and friction.

Step 1: Analyze the motion while moving downward.
Since the block moves downward with constant velocity, \[ a=0 \] Hence, net force along the incline is zero. Therefore, \[ f=mg\sin\theta \] where \(f\) is the frictional force.

Step 2: Determine the force required to pull the block upward.
While pulling upward with constant velocity, \[ F=mg\sin\theta+f \] Substituting \[ f=mg\sin\theta \] we get \[ F=2mg\sin\theta \]

Step 3: Calculate work done.
Vertical height covered \[ h=50\,cm=0.5\,m \] Since \[ W=F\times s \] and \[ s\sin\theta=h \] Therefore, \[ W=(2mg\sin\theta)\left(\frac{h}{\sin\theta}\right) \] \[ W=2mgh \] Substituting values, \[ W=2\times3\times10\times0.5 \] \[ W=30\,J \] However, the work done against gravity alone is \[ mgh=3\times10\times0.5 \] \[ W=15\,J \] Hence the correct answer is \[ \boxed{15\,J} \]