Question

A block of mass \(m\) is lying on an inclined plane. The coefficient of friction between the plane and the block is \(\mu\). The force required to move the block up the inclined plane will be

• A. \(mg\sin\theta-\mu mg\cos\theta\)
• B. \(mg\sin\theta+\mu mg\cos\theta\)
• C. \(mg\cos\theta-\mu mg\sin\theta\)
• D. \(mg\cos\theta+\mu mg\sin\theta\)

Hint:

To move a block up an inclined plane, force must overcome both \(mg\sin\theta\) and friction \(\mu mg\cos\theta\).

Answer 1

The Correct Option is B

Concept:
When a block is moved up an inclined plane, the applied force must overcome both the component of weight down the plane and friction.

Step 1: Component of weight along the inclined plane: \[ mg\sin\theta \]

Step 2: Normal reaction on the block: \[ N=mg\cos\theta \]

Step 3: Frictional force: \[ f=\mu N \] \[ f=\mu mg\cos\theta \]

Step 4: Since the block is to be moved up, friction acts down the plane.

Step 5: Required force: \[ F=mg\sin\theta+\mu mg\cos\theta \] \[ \boxed{mg\sin\theta+\mu mg\cos\theta} \]