Last 2 forces questions

  • Thread starter Cummings
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  • #1
Cummings
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http://cummingsiam.steven-sst.com/argh.jpg [Broken]

question 8 is similar to my last forces question. I got 3.2mm displacement for the raquet - yet the book says 2.2
The only difference i see is that the ball rebounds at 9.5ms-1 and instead of compression - we have displacement. Its the same thing...

question 9 is damnright confusing..
 
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  • #2
enigma
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How did you go about getting 3.2mm? Please post your work so we can check where your (or the book's) mistake is.

For the second one, remember Work = Force * distance. You'll need to do a few simple integrations.
 
  • #3
Cummings
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well,

the speed the ball is traveling is 10ms-1
the rule for kinetic energy is .5 * mass * velocity squared.
so .5 * .1 * 100
= 5j of energy

now, work done is measured in jouls.. and can be measured as the area under the force displacement graph.
so 5j = .5 * force * displacement.
if we let X = force and displacement
5j = .5Xsquared
10 = X squared
X = 3.2 mm displacement.

i let X = both force and displacement as if i change the scale of the force..X can equal both sides.

i get 3.2mm..the book says 2.2
 
  • #4
HallsofIvy
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"if we let X = force and displacement"????

I don't know what you mean by "scale the force". What scale are you using? That should depend on the "spring constant"- the slope of the Force versus displacement graph. Unfortunately, I was unable to open your link to find that. What is the slope of the graph?
You should be able to write the force as k*X, reading k from the graph.
 

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