A 0.50-kg block attached to an ideal spring with a spring constant of 80 N/m oscillates on a horizontal frictionless surface. The total mechanical energy is 0.12 J. The greatest extension of the spring from its equilibrium length is:

Question

Quizplus · Accepted Answer

The maximum extension of the spring can be found using the equation for total mechanical energy:
Total mechanical energy = 1/2*k*x^2 + 1/2*m*v^2
where k is the spring constant, x is the extension of the spring from its equilibrium position, m is the mass of the block, and v is the velocity of the block.

Since the surface is frictionless, there is no loss of mechanical energy due to friction. Therefore, the total mechanical energy is constant throughout the motion.

We are given that the total mechanical energy is 0.12 J. We can also find the velocity of the block at any point in the motion using conservation of energy:

Total mechanical energy = 1/2*k*x^2 + 1/2*m*v^2
0.12 J = 1/2*80 N/m*x^2 + 1/2*0.50 kg*v^2
v = sqrt((0.24 J - 40*x^2)/0.50 kg)

The maximum extension of the spring occurs when the block is at its maximum displacement from equilibrium, i.e., when the velocity is zero. Therefore, we can set v = 0 and solve for x:

0 = sqrt((0.24 J - 40*x^2)/0.50 kg)
0 = 0.24 J - 40*x^2
x = sqrt(0.24 J/40)/sqrt(0.50) = 0.054 m

Therefore, the greatest extension of the spring from its equilibrium length is 0.054 m. The answer is D.

A 050-Kg Block Attached to an Ideal Spring with a Spring