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3 years ago

It took 1700 J of work to stretch a spring from its natural length of 2m to a length of 5m. Find the spring's force constant.

2 answers:

7 0

The work to stretch a spring from its rest position is

   (1/2) (spring constant) (distance of the stretch)²

   E = 1/2 k x² .

You said it takes 1700 joules to stretch the spring 3 meters from its rest position, so we can write

   1700 joules = 1/2 k (3m)²

1 joule = 1 newton-meter

   1700 N-m = 1/2 k (3m)²

Multiply each side by 2: 3400 N-m = k · 9m²

Divide each side by 9m²    k = 3400 N-m / 9m²

   = (377 and 7/9)   newton per meter

Galina-37 [17]3 years ago

5 0

W = (1/2)kx^2, where W=work required to stretch the spring from its equilibrium position
k = spring constant
x = displacement = 5 - 2 = 3m
1800 J = (1/2)(k)(3)^2
by solving it we get
k = 400 answer.

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Hiii !!!
I am sending the soluction !!!

If you have any question, let me now =)

Jorge:)

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