Question

A 0.500 kg block, attached to a spring with length 0.60 m and force constant 40.0 N/m, is at rest with the back of the block at point A on a frictionless, horizontal air table (Figure 1). The mass of the spring is negligible. You move the block to the right along the surface by pulling with a constant 20.0 N horizontal force.

Answer 1

The velocity of the body at point B is approximately
`\( 3.873 \, \text{m/s} \)`

How to find the velocity of the body at point B?

Given:

Force applied,
`\( F = 20 \, \text{N} \)`

Displacement,
`\( x = 0.25 \, \text{m} \)`

Spring constant,
`\( k = 40.0 \, \text{N/m} \)`

According to the conservation of energy, the total energy at point B (sum of potential energy
`\( U_B \)` and kinetic energy
`\( K_B \)`) is equal to the work done:

`\[ U_B + K_B = W \]`

Where
`\( W = F * x = 20 * 0.25 = 5 \, \text{J} \)`

At point B:

`\[ K_B = (1)/(2)mv^2 \]`

`\[ U_B = (1)/(2)kx^2 \]`

So, combining these equations:

`\[ (1)/(2)kx^2 + (1)/(2)mv^2 = W \]`

`\[ (1)/(2) * 40 * (0.25)^2 + (1)/(2) * 0.5 * v^2 = 5 \]`

`\[ 2.5 + 0.5v^2 = 10 \]`

`\[ 0.5v^2 = 7.5 \]`

`\[ v = √(15) = 3.873 \, \text{m/s} \]`

Hence, the velocity of the body at point B is approximately
`\( 3.873 \, \text{m/s} \)`.

Complete question:

A 0.500 kg block is attached to a spring with length 0.60 m and force constant k=40.0 N/m. The mass of the spring is negligible. You pull the block to the right along the surface with a constant horizontal force F=20.0 N. What is the block's speed when the block reaches point B, which is 0.25 m to the right of point A?