7.1.30 The spring shown below has spring constant k. Its left end is fixed to th

Question

7.1.30 The spring shown below has spring constant k. Its left end is fixed to the wall. Its length is such that when it is at its resting length, its right end is at x = 0. Show that when a hand pulls the moving end of the spring to a position x, the x-component of the force exerted on the hand by the spring is F_u = -kx. Show that the work done on the hand by the spring as the hand stretches the spring from 0 to x is 1/2 kx^2. (You may do this either by integration, or by plotting F_u and finding the area under the curve. Do NOT use integration to solve the problem if you have not had a calculus class covering integration yet.)Assuming that the potential energy of an unstretched spring is 0, show that when the spring is stretched to x, the potential energy is 1/2kx^2.

Explanation / Answer

A)

by Newton's third law of motion

Action = -reaction

Force exerted on the spring by the hand = -Force exerted on the hand by the spring

by hooke's law Fsx = -k*x

b) work done by a variable force is W = integral (F*dx) = integral (-kx*dx) from o to x

W = -(1/2)*k*x^2 limits from 0 to x

W = (-1/2)*k*x^2

C) but work done is - change in potential energy

W = -(U2-U1) = U1-U2

-(1/2)*k*x^2 = 0 - U2

U2 = (1/2)*k*x^2...which is the required potential energy

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