Answer to Question #155495 in Physics for effa

Question

A mass on a spring system has a constant k=400 N/m and a mass m=0.50 kg.

(a) If it is given an initial displacement of 0.25 m and then released, what is the initial

potential energy of the spring?

(b) What is the maximum kinetic energy of the spring?

Answer 1

(a) By the definition of the elastic potential energy, we have:

$$PE=\dfrac{1}{2}kx^2,$$ $$PE=\dfrac{1}{2}\cdot 400\ \dfrac{N}{m}\cdot (0.25\ m)^2=12.5\ J.$$

(b) Let's first find the velocity of the mass when it passes through the equilibrium point:

$$PE=KE,$$ $$PE=\dfrac{1}{2}mv^2,$$ $$v=\sqrt{\dfrac{2PE}{m}},$$ $$v=\sqrt{\dfrac{2\cdot 12.5\ J}{0.50\ kg}}=7.1\ \dfrac{m}{s}.$$

Finally, we can calculate the maximum kinetic energy of the spring:

$$KE_{max}=\dfrac{1}{2}mv^2,$$ $$KE_{max}=\dfrac{1}{2}\cdot 0.50\ kg\cdot(7.1\ \dfrac{m}{s})^2=12.6\ J.$$

Answer:

(a) $PE=12.5\ J.$

(b) $KE_{max}=12.6\ J.$