You have two equal masses m1 and m2 and a spring with a spring constant k. The m

Question

You have two equal masses m1 and m2 and a spring with a spring constant k. The mass m1 is connected to the spring and placed on a frictionless horizontal surface at the relaxed position of the spring. You then hang mass m2, connected to mass m1 by a massless cord, over a pulley at the edge of the horizontal surface. When the entire system comes to rest in the equilibrium position, the spring is stretched an amount d1 as shown in figure (a). You are given the following information. The mass m1 = m2 = 0.490 kg. The spring constant k = 200 N/m. (a) Determine the amount the spring is stretched (d1) when m2 is attached to m1. d1 = m You now pull the mass m2 down a distance d2 = 6 cm and release it from rest, as shown in figure (b). Determine the following as the two masses travel the distance d2 back to their equilibrium positions. (The masses will overshoot the equilibrium position, but we are focusing our attention on them only as they travel the distance d2.) (b) Determine the work done on the system (m1, m2, and the massless connecting cord) by the spring. Ws = How is the work done by the spring related to the change in potential energy of the spring? J (c) Determine the work done on the system by the force of gravity. Wg = How is the work done by gravity related to the gravitational force and the vertical displacement of the masses? J (d) Determine the work done on the system by the normal force. WN = J (e) Determine the net work done on the system. Wnet = Since work is a scalar quantity, how may we determine the net work done? J (f) Determine the work done on m1 by the tension in the cord. WT1 = How is the change in kinetic energy of m1 related to the net work done on the system and the net work done on m1? What forces do work on m1? How is the change in kinetic energy of m1 related to the work done on m1 by the tension in the cord? J (g) Determine the work done on m2 by the tension in the cord.

Explanation / Answer

AS PER CHEGG'S POLICY, I WILL SOLVE FIRST FOUR SUB PARTS.....

(a) We know that

F = kx

so, x = F/k = 0.490*9.8 / 200 = 0.024 m

(b) Work done by spring = 1/2kx2 + mgh

Work done by spring = 1/2*200*0.062 + 0.490*9.8*0.06

Work done by spring = 0.648 J

(c) Work done by force of gravity = - mgh

  Work done by force of gravity = -0.49*9.8*0.06

  Work done by force of gravity = - 0.288 J

(d) Work done by normal force = 0 J

(e) Net work done = 0.648 + (-0.288) + 0

Net work done = 0.36 J

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