Question

8. A 0.20-kg object is oscillating on a spring with a spring constant of k = 15 Nm.
What is the potential energy of the system when the object displacement is 0.040 m.
exactly half the maximum amplitude?
a. zero
b. 0.006 0
d. 25)
​

Answer 1

Explanation:

Refer to the attachment!

Answer 2

c. The potential energy of the system when the object displacement is 0.040m, exactly half the maximum amplitude is 0.012J.

Given,

k = 15Nm.

x = 0.040m.

To find,

Potential energy when object displacement is at half the maximum amplitude.

Formula used,

P.E = $\frac{1}{2} kx^{2}$

• Potential energy is energy possessed by any particle when at rest.
• The total energy is generally the total of kinetic as well as potential energies of the particle.
• Whenever kinetic energy is maximum potential energy tends to be zero, and vice-versa.

P.E = $\frac{1}{2} kx^{2}$

By substituting values,

$P.E = \frac{1}{2}[15][0.040][0.040]\\ \\P.E = 0.012Nm^{2}$

Since, N$m^{2}$ = Joules

Therefore, the potential energy of the system when the object displacement is 0.040m, exactly half the maximum amplitude is 0.012J.