Question

A box of mass 4.3 kg is placed against a spring of spring constant k=97 N/m. The spring is then compressed a distance Δx=0.48 m from its equilibrium (unstretched, uncompressed) length and the box is held stationary against the spring. The box is resting on a table, and the coefficient of kinetic friction between the box and the table is μₖ=0.18. If the box is released, how far will the block travel before stopping?

Answer 1

The block will travel a distance of approximately 0.077 m before stopping.

Step-by-step explanation:

To determine the distance the block will travel before stopping, we need to consider the forces acting on the block. First, we can calculate the force exerted by the spring using Hooke's Law:

F = kΔx

where F is the force, k is the spring constant, and Δx is the displacement from equilibrium. Plugging in the given values:

F = (97 N/m)(0.48 m) = 46.56 N

Next, we calculate the force of kinetic friction using the coefficient of kinetic friction:

fₖ = μₖmg

where fₖ is the force of kinetic friction, μₖ is the coefficient of kinetic friction, m is the mass of the block, and g is the acceleration due to gravity. Plugging in the given values:

fₖ = (0.18)(4.3 kg)(9.8 m/s²) = 7.3206 N

Since the block is held stationary against the spring, the forces of the spring and kinetic friction must balance:

F = fₖ

Solving for the displacement Δx:

kΔx = μₖmg

Δx = (μₖmg) / k

Plugging in the values:

Δx = [(0.18)(4.3 kg)(9.8 m/s²)] / (97 N/m) = 0.077 m

Therefore, the block will travel a distance of approximately 0.077 m before stopping.