stored spring energy


by lemon
Tags: energy, spring, stored
lemon
lemon is offline
#1
Apr22-10, 11:06 AM
P: 203
Hi - Would someone check my method here please?
Thank you

1. The problem statement, all variables and given/known data

Two trolleys A and B, of mass 0.70kg and 0.80kg respectively, are on a horizontal track and held together by a compressed spring. When the spring is released the trolleys separate freely and A moves to the left with an initial velocity of 4.0m/s.
Calculate:
a) the initial velocity of A
b) the energy initially stored in the spring



2. Relevant equations

F=ma



3. The attempt at a solution

a) (0.70x4.0)=(0.80xV)
v=(0.70x4.0)/(0.80)
v=3.5m/s

b) This is a transfer of elastic potential energy to kinetic energy -
Energy=1/2mv^2
mv=p(momentum)
1/2(0.70x4.0)^2 + 1/2(0.80x3.5)^2
=0.0392 + 3.92
=3.9592 J
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Doc Al
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#2
Apr22-10, 11:34 AM
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Quote Quote by lemon View Post
a) (0.70x4.0)=(0.80xV)
v=(0.70x4.0)/(0.80)
v=3.5m/s
This is good. You used conservation of momentum.

b) This is a transfer of elastic potential energy to kinetic energy -
Energy=1/2mv^2
mv=p(momentum)
1/2(0.70x4.0)^2 + 1/2(0.80x3.5)^2
=0.0392 + 3.92
=3.9592 J
Your mistake here was squaring mv, instead of just v.
KE = 1/2mv^2, but you calculated 1/2(mv)^2. Not the same!
lemon
lemon is offline
#3
Apr22-10, 11:40 AM
P: 203
ahh of course:
silly boy!

(1/2x0.7x4.0^2) + (1/2x0.8x3.5^2) = 10.5 J

Doc Al
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#4
Apr22-10, 11:41 AM
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P: 40,877

stored spring energy


Now you've got it.
lemon
lemon is offline
#5
Apr22-10, 12:44 PM
P: 203
Thanks Doc


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