A block of mass ‘m’ is placed on a rough inclined plane When the inclination of the plane is θ, the block just begins to slide down the plane under its own weight. The minimum force applied parallel to the plane, to move the block up the plane, is

Question

A block of mass ‘m’ is placed on a rough inclined plane When the inclination of the plane is θ, the block just begins to slide down the plane under its own weight. The minimum force applied parallel to the plane, to move the block up the plane, is (A) mg sin θ (B) 2mg sin θ (C) mg cos θ (D) mg tan θ

Tardigrade Admin · Accepted Answer

As object just begin to slide f =mg sin θ .....(i) If we start sliding the block up, the friction force will be downward direction, hence minimum force applied in upward direction <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/525280b9f8aac5c25cfe71b607899b11-.png" /> Component of weight along inclined plane + force of friction =2mg sin θ

Q. A block of mass ‘m’ is placed on a rough inclined plane When the inclination of the plane is $θ$ , the block just begins to slide down the plane under its own weight. The minimum force applied parallel to the plane, to move the block up the plane, is

$m g s in θ$

$2 m g s in θ$

$m g cos θ$

$m g t an θ$

As object just begin to slide

$f = m g s in θ ..... (i)$

If we start sliding the block up, the friction force will be downward direction, hence minimum force applied in upward direction

Component of weight along inclined plane + force of friction

$= 2 m g s in θ$

Q. A block of mass ‘m’ is placed on a rough inclined plane When the inclination of the plane is θ, the block just begins to slide down the plane under its own weight. The minimum force applied parallel to the plane, to move the block up the plane, is

mgsinθ

2mgsinθ

mgcosθ

mgtanθ