no idea all HELP appreciated!!
A shot-putter puts a shot (weight = 70.3 N) that leaves his hand at distance of 1.45 m above the ground.
(a) Find the work done by the gravitation force when the shot has risen to a height of 2.15 m above the ground. Include the correct sign for work.
(b) Determine the change (ÄPE = PEf - PE0) in the gravitational potential energy of the shot.
To find the work done by the gravitational force when the shot has risen to a height of 2.15 m above the ground, we can use the formula for work:
Work = Force x Distance x Cos(angle)
(a) Using the given information, we know that the weight of the shot (force) is 70.3 N and the distance it has risen is 1.45 m. The angle between the direction of the force and the displacement is 180 degrees (since the shot is moving upward against gravity). Therefore, the formula becomes:
Work = 70.3 N x 1.45 m x Cos(180°)
The cosine of 180 degrees is -1, so the formula simplifies to:
Work = -70.3 N x 1.45 m x (-1)
Simplifying further, we get:
Work = 70.3 N x 1.45 m
So, the work done by the gravitational force is 101.935 J (joules). The correct sign for work is negative, indicating that the force opposes the displacement.
(b) To determine the change in gravitational potential energy (ΔPE) of the shot, we can use the formula:
ΔPE = mgh
Where ΔPE is the change in potential energy, m is the mass of the shot, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the change in height.
Since the weight of the shot is given as 70.3 N, which is equal to the force of gravity (mg), we can assume the mass of the shot is 7.16 kg (70.3 N divided by 9.8 m/s^2).
Given that the shot has risen from a height of 1.45 m to 2.15 m, the change in height is 0.7 m. Substituting the values into the formula, we get:
ΔPE = 7.16 kg x 9.8 m/s^2 x 0.7 m
Simplifying further, we find that the change in gravitational potential energy is 48.59 J (joules).