Question

a man of mass m climbs up a hanging vertical rope for a distance d. work done, W, by friction force acting on his hands is

Answer 1

Answer:

man and the rope, the man will slide down the ground. It is the frictional force which take him in upward direction.

So,

F−mg=ma

F=mg+ma

F=m(g+a)↑

Answer 2

A man of mass m climbs up a hanging vertical rope for a distance d. Work done, W, by friction force acting on his hands is (md(g+a)).

• Two kinds of forces are acting on the man.
• One force is acting downwards which is the product of his mass and the acceleration due to gravity.

        $F_{1} = mg$

• The other force is the frictional force which is acting upwards.

        $F_{2} = f$

• These forces are equal to the product of the mass and the acceleration of the man.

       $f - mg = ma$

       $f = mg + ma$

          $= m (g+a)$

• Work done by a body is the product of the force applied to it and its displacement.

        $W = f * d$

• Here, W is the work done, f is the frictional force acting and d is the distance.

        $W = m(g+a) * d$

             $= md(g+a)$