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The potential energy of a spring is given by the equation PE

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The potential energy of a spring is given by the equation PE

by alanforde800Maximus » Tue Oct 18, 2016 8:25 am
The potential energy of a spring is given by the equation PE = (kx^2)/2 where k is a constant and x is the distance the spring is stretched. If K =16 and the spring is stretched to 2 feet and then to 3 feet, how much potential energy is gained by the spring from the moment it is stretched 2 feet to the moment it is stretched 3 feet?

a) 200
b) 128
c) 72
d) 40
e) 32

Please assist with above problem.
Last edited by alanforde800Maximus on Tue Oct 18, 2016 9:08 pm, edited 1 time in total.
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by regor60 » Tue Oct 18, 2016 9:14 am
alanforde800Maximus wrote:The potential energy of a spring is given by the equation PE = (kx^2)/2 where k is a constant and x is the distance the spring is stretched. If x =16 and the spring is stretched to 2 feet and then to 3 feet, how much potential energy is gained by the spring from the moment it is stretched 2 feet to the moment it is stretched 3 feet?

a) 200
b) 128
c) 72
d) 40
e) 32

Please assist with above problem.
I think you mean K=16, otherwise it's not solvable.

Assuming this is the case, going from 2 feet to 3 feet in the equation is 16/2(3^2 - 2^2) = 8*(9-4) = D
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by alanforde800Maximus » Tue Oct 18, 2016 9:09 pm
regor60 wrote:
alanforde800Maximus wrote:The potential energy of a spring is given by the equation PE = (kx^2)/2 where k is a constant and x is the distance the spring is stretched. If x =16 and the spring is stretched to 2 feet and then to 3 feet, how much potential energy is gained by the spring from the moment it is stretched 2 feet to the moment it is stretched 3 feet?

a) 200
b) 128
c) 72
d) 40
e) 32

Please assist with above problem.
I think you mean K=16, otherwise it's not solvable.

Assuming this is the case, going from 2 feet to 3 feet in the equation is 16/2(3^2 - 2^2) = 8*(9-4) = D

yes, it's K =16 sorry for typo.
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