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A proud deep-sea fisherman hangs a 69.5-kg fish from an ideal spring having negligible mass. The fish stretches the spring 0.130 m. (a) Find the force constant of the spring.

Respuesta :

Force constant = k = 6811 N/m

Given:

mass = m = 69.5 kg

L(initial) = ideal spring length = 12 cm= 0.120 m

L(final) = final spring length = 13 cm= 0.130 m

Two forces acting on the object is k*x and mg

so,

kx-mg=0

kx = mg

k = mg/x

We know that x= displacement = L(final)- L(initial)

0.130-0.120=0.10

k= 69.5×9.8/0.10

k =6811 N/m

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