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Work of 44 joules is done in stretching a spring from its natural length to 18 cmcm beyond its natural length. what is the force (in newtons) that holds the spring stretched at the same distance (18 cmcm)?
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Work of 44 joules is done in stretching a spring from its natural length to 18 cmcm beyond its natural length. what is the force (in newtons) that holds the spring stretched at the same distance (18 cmcm)?

User Miboper
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1 Answer

6 votes
When springs are stretch, they store potential energy. The equation can be modeled by 0.5kx^2

Substituting known values, we get


44=0.5k(0.18)^(2) k=2716.049N/m

Next use Hooke's law F=-kx to solve for the force.


F=-(2716.049N/m)(0.18m) F=488.88889N
User DanielMason
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