Physics Review Question

Unit: WORK AND ENERGY
Year: 1995 Question#: 17
Question: When a spring is stretched 0.200 meter from its equilibrium position, it possesses a potential energy of 10.0 joules. What is the spring constant for this spring?

(1 ) 100. N/m
(2 ) 125 N/m
(3 ) 250. N/m
(4 ) 500. N/m

What is this question really asking?




Explanation by: JD W.

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Answer 1
100. N/m

INCORRECT: The equation for this spring constant problem is PE=1/2kx2. If you use this equation with the given information, the answer is not 100 N/m. return to top























Answer 2
125 N/m

INCORRECT

INCORRECT: The equation for this spring constant problem is PE=1/2kx2.
return to top



























Answer 3
250. N/m


INCORRECT
The equation for this spring constant problem is PE=1/2kx2. return to top


























Answer 4
500. N/m


CORRECT

Answer 4 is correct. the equation is PE=1/2kx2. SO:
GIVEN: PE=10J
x=0.2m

PE=1/2kx2
10J=1/2(k)(0.2m)2
10J=0.02m2k
k=500N/m return to top


























What's this question really asking?

This question is really asking what the spring constant is. The spring constant is the ratio of the force required to stretch or compress a particular spring to the distance of stretch or compression. If you know the spring constant you can then apply the formula PE=1/2kx2. return to top

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