A force of magnitude 800 N stretches a certain spring by 0.200 m from its equili

Question

A force of magnitude 800 N stretches a certain spring by 0.200 m from its equilibrium position.

Part A

What is the force constant of this spring?

k = _______ N/m

Part B

How much elastic potential energy is stored in the spring when it is stretched 0.300 m from its equilibrium position.

U = _______ J

Part C

How much elastic potential energy is stored in the spring when it is compressed by 0.300 m from its equilibrium position?

Part D

How much work was done in stretching the spring by the original 0.200 m ?

Explanation / Answer

A) F = kx

800 = k (0.200)

k = 4000 N /m

b) U = kx^2 /2

U = 4000 x 0.3^2 /2 = 180 J

c) U = kx^2 /2

U = 4000 x 0.3^2 /2 = 180 J

D) acting force is constant = 800 N

work done = F.d = 800 x 0.200 = 160 J

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