A 4.55 kg steel ball is dropped onto a copper plate from a height of 10.0 m. If the ball leaves a dent 2.80 mm deep in the plate, what is the average force exerted by the plate on the ball during the impact?
To find the average force exerted by the plate on the ball during the impact, we can use the principles of conservation of energy. The potential energy of the ball at the initial height is converted into kinetic energy as it falls, and then into work done on the ball as it comes to a stop and creates a dent in the plate.
1. Calculate the gravitational potential energy of the ball at the initial height:
Potential energy (PE) = mass (m) x gravity (g) x height (h)
PE = 4.55 kg x 9.8 m/s² x 10.0 m
2. Calculate the work done on the ball to create the dent in the plate:
Work (W) = force (F) x distance (d)
In this case, distance is the depth of the dent created by the ball.
Work = F x 2.80 mm
3. Equate the potential energy to the work done on the ball:
PE = W
4.55 kg x 9.8 m/s² x 10.0 m = F x 2.80 mm
4. Convert millimeters to meters:
2.80 mm = 2.80 x 10⁻³ m
5. Solve for the force (F):
F = (4.55 kg x 9.8 m/s² x 10.0 m) / (2.80 x 10⁻³ m)
F = 1.75 x 10⁵ N
Therefore, the average force exerted by the plate on the ball during the impact is approximately 1.75 x 10⁵ Newtons.