Question

A shot-putter puts a shot (weight = 71.5 N) that leaves his hand at a distance of 1.84 m above the ground. (a) Find the work done by the gravitational force when the shot has risen to a height of 2.50 m above the ground. Include the correct sign for the work. (b) Determine the change (PEf - PE0) in the gravitational potential energy of the shot.

Answer 1

Answer:

a)W= - 47.19 J

b)47.19 J

Explanation:

Given that

Weight ,mg= 71.5 N

y= 1.84 m

H= 2.5

a)

The distance above the hand ,h= 2.5 - 1.84 m

   h= 0.66 m

We know that gravitaional force act in the downward direction but the displacement is in upward direction that is why work done will be negative.

W= - m g h

W= -71.5 x 0.66 J

W= - 47.19 J

b)

The potential energy at initial position = m g y

The potential energy at final position = m g H

So change in the potential energy = m g H - m g y

                                                           = mg ( H- y)

                                                           =71.5 ( 2.5 - 1.84 ) = 47.19 J

Therefore change in the potential energy = 47.19 J