A spring (k = 600 N/m) is at the bottom of a frictionless plane that makes an angle of 30° with the horizontal. The upper end of the spring is depressed 0.10 m, and a 2.0-kg block is placed against the depressed spring. The system is then released from rest. What is the kinetic energy of the block at the instant it has traveled 0.10 m and the spring has returned to its uncompressed length?
A) 2.0 J
B) 1.8 J
C) 2.2 J
D) 1.6 J
E) 1.0 J

Question

Quizplus · Accepted Answer

The potential energy stored in the spring when compressed is (1/2)kx^2 = (1/2)(600 N/m)(0.10 m)^2 = 3.0 J. As the spring returns to its uncompressed length, this energy is converted into kinetic energy and gravitational potential energy. The gravitational potential energy change is mgh = (2.0 kg)(9.8 m/s^2)(0.10 m * sin(30°)) = 0.98 J. Therefore, the kinetic energy is 3.0 J - 0.98 J = 2.02 J, which is approximately 2.0 J.

A Spring (K = 600 N/m) Is at the Bottom