A certain spring is found not to conform to Hooke\'s law. The force (in newtons)

Question

A certain spring is found not to conform to Hooke's law. The force (in newtons) it exerts when stretched a distance x (in meters) is found to have magnitude 42.7x + 25.3x2 in the direction opposing the stretch. (a) Compute the work required to stretch the spring from x = 0.32 m to x = 0.64 m. (b) With one end of the spring fixed, a particle of mass 2.46 kg is attached to the other end of the spring when it is extended by an amount x = 0.64 m. If the particle is then released from rest, what is its speed at the instant the stretch in the spring is x = 0.32 m? (c) Is the force exerted by the spring conservative or nonconservative?

Explanation / Answer

a) F = 42.7x + 25.3x^2

Work done dW = Fdx = (42.7x + 25.3x^2)dx

intergrating both side

W = 21.35x^2 + 8.43x^3

work done from x = 0.32m to 0.64 m

W = 21.35 ( 0.64^2 - 0.32^2 ) + 8.43 (0.64^3 - 0.32^3 )

W = 8.49 J

b) Work done = change in KE

8.49 = 2.46v^2 /2 - 0

v = 2.63 m/s

c) this is conservative force

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