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

AI Thread Summary
A 4.00 kg steel ball dropped from 10.0 m impacts a copper plate, creating a dent of 2.75 mm. The kinetic energy at impact is calculated to be 76,832 J, assuming no air resistance. The work done on the plate is equal to the force exerted multiplied by the dent depth. By equating gravitational potential energy to the work done, the average force exerted by the plate is derived. The calculations suggest a need for reevaluation, as initial results appear excessively high.
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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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