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 42.5x + 47.5x2 in the direction opposing the stretch. (a) Compute the work required to stretch the spring from x = 0.47 m to x = 0.94 m. (b) With one end of the spring fixed, a particle of mass 3.60 kg is attached to the other end of the spring when it is extended by an amount x = 0.94 m. If the particle is then released from rest, what is its speed at the instant the stretch in the spring is x = 0.47 m?

Explanation / Answer

work done=F*dx

F=K*x

W=integration from 0.47 to 0.94( 42.5x + 47.5x2)dx

=25.59 J

b.)when x=0.47 m

change in potential energy of the spring=change in kinetik energy

1/2*3.6*v^2=25.59

v=3.77 m/s ans

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