A shot putter puts a shot (weight=70.3N) that leaveshis hand at distance of 1.51m above the ground.
(a) Find the work done by the gravitatiob force when the shot has risen to a height of 2.10m above the ground. include the correct sign for work.
(b) Determine the change in the gravitational potential energy of the shot.
a) Since work is done against the gravitational force, the sign is negative. The magnitude is
M g (delta h),
where "delta h" is the height change, 0.59 m, gi is the acceleration of gravity and M is the mass.
b) This would be the same nummber with opposite (positive) sign.
To find the work done by the gravitational force when the shot has risen to a height of 2.10m, you need to calculate the negative of the product of the weight (70.3N) and the height change (2.10m - 1.51m = 0.59m).
a) Work = -M * g * delta h
Substituting the given values:
Work = -70.3N * 9.8m/s^2 * 0.59m
Work ≈ -408.02 Joules
Therefore, the work done by the gravitational force is approximately -408.02 Joules. The negative sign indicates that work is done against the gravitational force.
To determine the change in gravitational potential energy of the shot, you can use the same magnitude of work but with the opposite (positive) sign:
b) Change in gravitational potential energy = -Work
Substituting the calculated value from part (a):
Change in gravitational potential energy ≈ 408.02 Joules