Calculating Work Done by Spring at Various Positions

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The discussion revolves around calculating the work done by a spring as a block moves through various positions. The spring constant (k) is calculated using the formula k = Fs/d, resulting in a value of 1428.57 N/m based on a force of 200 N at a displacement of 14 cm. However, confusion arises regarding the interpretation of the problem's wording, particularly the phrase "the scale is set by Fs = 200 N." Clarification is sought on whether this means the spring force is 200 N when stretched to 14 cm. Ultimately, the original poster indicates they have solved the problem, resolving their initial confusion.
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Homework Statement


The scale is set by Fs = 200 N. We release the block at x = 14.0 cm. How much work does the spring do on the block when the block moves from xi =+8.0 cm to (a)x=+7.0 cm, (b)x=-7.0 cm, (c)x=-8.0 cm, and (d)x=-11.0 cm?
xi= initial point
x= final point

Homework Equations





The Attempt at a Solution


I'm required to obtain K first before proceeding to the next step.
Fs= -kd
k= - Fs/d
= 200/0.14
= 1428.57 N/m

K= spring constant
d=distance
But the answer is wrong. May I know the reason ? Thank you.
 
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The wording of the problem is unclear. If it means that when the spring is stretched 14 cm, the force in the spring is 200 N, then your answer is correct. But what is really meant by "the scale is set by Fs = 200 N"?
 
PhanthomJay said:
The wording of the problem is unclear. If it means that when the spring is stretched 14 cm, the force in the spring is 200 N, then your answer is correct. But what is really meant by "the scale is set by Fs = 200 N"?
Thank you very much. I solved this question already. Love you.
 
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