What is the average force exerted by the plate on the ball

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SUMMARY

The average force exerted by a copper plate on a 4.00 kg steel ball dropped from a height of 10.0 m, resulting in a dent of 2.75 mm, is calculated using energy conservation principles. The kinetic energy (KE) at impact is 76832 J, derived from the equation KE = 1/2 mv². The work done (W) during the impact, represented as W = Fd, leads to the equation mgh = Fs, where F is the average force. The calculated average force is approximately 27938909.1 N, indicating a significant discrepancy in the initial assumptions regarding energy conversion and impact dynamics.

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  • Understanding of Newton's laws of motion
  • Familiarity with kinetic energy and gravitational potential energy concepts
  • Knowledge of work-energy principle
  • Basic algebra for solving equations
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  • Study the work-energy theorem in detail
  • Learn about impact forces and material deformation
  • Explore energy conservation in mechanical systems
  • Investigate the effects of air resistance on falling objects
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Physics students, mechanical engineers, and anyone interested in understanding impact forces and energy transfer in collisions.

mkwok
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Homework Statement


A 4.00 kg steel ball is dropped onto a copper plate from a height of 10.0 m. If the ball leaves a dent 2.75 mm deep, what is the average force exerted by the plate on the ball during the impact?


Homework Equations


KE=1/2mv^2
F=ma
Wgrav=mgh


The Attempt at a Solution



when the ball hit the plate, the ball is traveling at a speed of 2ax, which is also 196m/s
so the Kinetic Energy would be 76832 the moment the ball hits the plate

since W=Fd, so 76832=F(2.75/1000)
F=27938909.1N

just looking by the numbers, I can tell this is wrong
 
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Ignoring air resistance..all the gravitational potential energy is converted into kinetic energy as the mass falls...when the ball hits the copper plate, the work done in moving the ball 2.75mm into the copper. so that basically the loss in k.e. is equal to the work done in moving it into the copper
W=Fs so that mgh=Fs
 

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