A spring,with k = 2500N/m,is fixed vertically to the floor.A 10.0-kg box is then pressed down such that the spring is compressed 0.5 m beyond its equilibrium length.The box is then released.Assuming the work done against drag (air resistance)is 60 J,how high will the box travel above its release position?

Question

Quizplus · Accepted Answer

Explanation:The potential energy stored in the spring is:$$U = \frac{1}{2}kx^2 = \frac{1}{2}(2500 N/m)(0.5 m)^2 = 625 J$$The work done against drag is 60 J, so the kinetic energy of the box when it is released is:$$K = U - W = 625 J - 60 J = 565 J$$The maximum height the box will travel above its release position is:$$h = \frac{K}{mg} = \frac{565 J}{(10 kg)(9.8 m/s^2)} = 5.76 m$$However, the box will not actually reach this height because of air resistance. The actual height it will reach is:$$h' = h - \frac{W}{mg} = 5.76 m - \frac{60 J}{(10 kg)(9.8 m/s^2)} = 2.58 m$$Therefore, the correct answer is A.

A Spring,with K = 2500N/m,is Fixed Vertically to the Floor