Complete 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:
The kinetic energy is [tex]KE = 0.4368\ J[/tex]
Explanation:
From the question we are told that
The mass of the block is [tex]m= 0.025\ kg[/tex]
The spring constant is [tex]k = 150 N/m[/tex]
The length of first displacement is [tex]x_1 = 0.80 \ m[/tex]
The length of first displacement is [tex]x_2 = 0.024 \ m[/tex]
At the [tex]x_2[/tex] the kinetic energy is mathematically evaluated as
[tex]KE = \Delta E[/tex]
Where [tex]\Delta E[/tex] is the change in energy stored on the spring which is mathematically represented as
[tex]\Delta E = \frac{1}{2} k (x_1 ^2 - x_2^2)[/tex]
=> [tex]KE = \frac{1}{2} k (x_1 ^2 - x_2^2)[/tex]
Substituting value
[tex]KE = \frac{1}{2} * 150 * (0.08^2 - 0.024^2)[/tex]
[tex]KE = 0.4368\ J[/tex]
