A spring (k = 600 N/m)is at the bottom of a frictionless plane that makes an angle of 30$\degree$ 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 of the block. The gravitational potential energy change is mgh = (2.0 kg)(9.8 m/s^2)(0.10 m * sin(30°)) = 0.98 J. Thus, the kinetic energy is 3.0 J - 0.98 J = 2.02 J, approximately 2.0 J.

A Spring (K = 600 N/m)is at the Bottom of a Frictionless