Question

A 0.50-kg mass is attached to a spring of spring constant 20 N/m along a horizontal, frictionless surface. The object oscillates in simple harmonic motion and has a speed of 1.5 m/s at the equilibrium position. What is the total energy of the system?

Answer 1

Hi there!

With the work-energy theorem for oscillating springs:

ME = KE + PE

$ME = \frac{1}{2}mv^2 + \frac{1}{2}kx^2$

Where:

m = mass (kg)

v = velocity (m/s)

k = Spring Constant (N/m)

x = displacement from equilibrium (m)

If the object is at the equilibrium position, there is NO potential energy since:

$\frac{1}{2}k(0^2) = 0 J$

Thus:

$ME = \frac{1}{2}mv^2$

Plug in the given values:

$ME = \frac{1}{2}(0.50)(1.5^2) = \boxed{0.5625 \text{ J}}$