Question

A 0.025-kg block on a horizontal frictionless surface is attached to an ideal massless spring whose spring constant is 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 what is the kinetic energy of the block?

Answer 1

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
`KE = 0.4368\ J`

Step-by-step explanation:

From the question we are told that

The mass of the block is
`m= 0.025\ kg`

The spring constant is
`k = 150 N/m`

The length of first displacement is
`x_1 = 0.80 \ m`

The length of first displacement is
`x_2 = 0.024 \ m`

At the
`x_2` the kinetic energy is mathematically evaluated as

`KE = \Delta E`

Where
`\Delta E` is the change in energy stored on the spring which is mathematically represented as

`\Delta E = (1)/(2) k (x_1 ^2 - x_2^2)`

=>
`KE = (1)/(2) k (x_1 ^2 - x_2^2)`

Substituting value

`KE = (1)/(2) * 150 * (0.08^2 - 0.024^2)`

`KE = 0.4368\ J`