A 2.50-kg object is attached to a spring of force constant k = 4.50 kN/m. The spring is stretched 10.0 cm from equilibrium and released. What is the maximum kinetic energy of this system?
A) 45.0 J
B) 22.5 J
C) 56.0 J
D) 2.25 * 105 J
E) 4.50 J

Question

Quizplus · Accepted Answer

The maximum kinetic energy occurs when all of the potential energy stored in the spring is converted to kinetic energy.

The potential energy stored in the spring when it is stretched x meters is given by:

PE = (1/2)kx^2

In this case, the spring is stretched 10 cm = 0.10 m from equilibrium, so the potential energy stored in the spring is:

PE = (1/2)(4.50 kN/m)(0.10 m)^2 = 0.0225 kJ = 22.5 J

At the maximum kinetic energy point, all of this potential energy has been converted to kinetic energy, so:

KE = 22.5 J

Therefore, the answer is choice B.

A 250-Kg Object Is Attached to a Spring of Force Constant