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, applied force must overcome both gravity component \(mg\sin\theta\) and friction \(\mu mg\cos\theta\).

Answer 1

The Correct Option is B

A block is on an inclined plane of angle \(\theta\). The weight of the block is: \[ mg. \] The component of weight along the inclined plane downward is: \[ mg\sin\theta. \] The normal reaction is: \[ N=mg\cos\theta. \] The frictional force is: \[ f=\mu N. \] Substitute \(N=mg\cos\theta\): \[ f=\mu mg\cos\theta. \] Now, the block is to be moved up the inclined plane. When the block moves upward, friction acts downward along the plane. So the applied force must overcome two downward forces: \[ mg\sin\theta \] and \[ \mu mg\cos\theta. \] Therefore, required force is: \[ F=mg\sin\theta+\mu mg\cos\theta. \] Hence, the force required is: \[ mg\sin\theta+\mu mg\cos\theta. \]