Question

A 0.025-kg block on a horizontal frictionless surface is attached to an ideal massless spring whose spring constant is 150 N/m. The block is pulled from its equilibrium position at x = 0.00 m to a displacement x = +0.080 m and is released from rest. The block then executes simple harmonic motion along the horizontal x-axis. When the displacement is x = 0.024 m, what is the kinetic energy of the block?

Answer 1

Answer:

K=0.4368J

Explanation:

The mechanical energy will be the sum of the elastic energy and the kinetic energy of the block: $E=U+K$. Their formulas are $U=\frac{k\Delta x^2}{2}$ and $K=\frac{mv^2}{2}$ for any point in the movement (although E must remain constant since the movement is frictionless), where k is the spring constant, $\Delta x$ the displacement, m the mass of the block and v its velocity.

When the block is released from rest, at that point its mechanical energy is equal to the elastic energy, which is at its maximum ($E=U_{max}$) since the displacement is at its maximum ($\Delta x_{max}=0.08m$).

Since mechanical energy is conserved, the mechanical energy at that point must be equal also to the mechanical energy at the point asked, which we will call P (when the displacement is $x_P=0.024m$), so we will have:

$E=U_{max}=U_P+K_P$

So the kinetic energy at point P is

$K_P=U_{max}-U_P=\frac{k \Delta x_{max}^2}{2}-\frac{k \Delta x_P^2}{2}=\frac{k (\Delta x_{max}^2-\Delta x_P^2)}{2}$

And substituting values we have:

$K_P=\frac{(150N/m) ((0.08m)^2-(0.024m)^2)}{2}=0.4368J$