Question

A shot-putter puts a shot (weight = 71.9 N) that leaves his hand at distance of 1.47 m above the ground.
(a) Find the work done by the gravitation force when the shot has risen to a height of 2.20 m above the ground. Include the correct sign for work.
J
(b) Determine the change (ÄPE = PEf - PE0) in the gravitational potential energy of the shot.
J

Answer 1

The shot rises against gravity for (2.20 - 1.47) = .73 m
therefore the work done which is negative because the force is down and the motion up = - m g (.73) = -71.9(.73) = -52.5 Joules

b. 52.5 Joules
comes out of kinetic energy and goes to potential energy