Last 2 forces questions

Cummings

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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enigma

Staff Emeritus
Gold Member
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.

Cummings

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

HallsofIvy

Homework Helper
"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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