Question

A external agent moves the block m slowly from A to B,
along a smooth hill such that every time he applies theforce tangentially. Find the work done by agent in thisinterval

Answer 1

Apply work energy theorem and equate it to 0
Bcoz the blocked is moved slowly...i.e change in KE=0
Mgh+work done by external agent=0

Answer 2

The work done by agent in this interval is  Wagent = mgh

1. Using work energy theorem
2. Change in the kinetic energy = total work done
3. In the given case the change in the kinetic energy is zero
4. 0= work done by gravity (Wg) + work done by agent/external force (Wagent)
5. The work done by gravity is = -mgh
6. Here m mass of the block , g is the acceleration due to gravity and h is the height
7. 0 = -mgh + Wagent
8. The work done by agent in this interval Wagnet = mgh