Question

A 5-kilogram block is fastened to a vertical spring that has a spring constant of 1,000 newtons per meter. A

3-kilogram block rests on top of the 5-kilogram block, as shown above. When the blocks are at rest, how much is the spring compressed from its original length?

Answer 1

To find the spring compression at rest, the gravitational force of both blocks is equated to the spring's elastic force, resulting in a compression of 7.8 cm.

Step-by-step explanation:

To determine how much the spring is compressed when the blocks are at rest, we need to consider the force exerted by the blocks due to gravity and match that with the elastic force of the spring given by Hooke's law (F = kx).

The total weight of the blocks is the force due to gravity acting on them.

This force is calculated by multiplying the total mass of the blocks (5 kg + 3 kg) by the acceleration due to gravity (g = 9.8 m/s²). So, the force is $F_{gravity} = (5 kg + 3 kg) × 9.8 m/s² = 78 N$.

Hooke's law for the spring force is $F_{spring} = kx$, where k is the spring constant and x is the displacement of the spring from its equilibrium position.

For equilibrium (when the blocks are at rest), the spring force must equal the gravitational force on the blocks: $F_{spring} = F_{gravity}$.

So, $1000 N/m × x = 78 N$. Solving for x, we get $x = 78 N / 1000 N/m = 0.078 m$, which is 7.8 cm.

Therefore, the spring is compressed 7.8 cm from its original length when the blocks are at rest.