A box is set into motion from rest and comes to rest in 5.0 s. If the coefficient of kineticfriction is 0.45, what was the box’s initial velocity?

1. Break Down the Problem

We need to determine the initial velocity of a box that comes to rest after 5.0 seconds, while experiencing kinetic friction. The steps to find the solution include:

1. Analyze the forces acting on the box.
2. Apply the equations of motion and friction to express the relationship between initial velocity, final velocity, and acceleration.
3. Solve for the initial velocity.

2. Relevant Concepts

• The box starts from rest, which means its final velocity, $v_f$, is 0 m/s.
• The coefficient of kinetic friction $\mu_k$ is 0.45.
• The acceleration $a$ can be determined from the frictional force.
• Using Newton’s second law: $F_{\text{friction}} = \mu_k \cdot F_{\text{normal}} = \mu_k \cdot m \cdot g$
• Apply kinematic equations: $v_f = v_i + a \cdot t$

3. Analysis and Detail

1. Calculate the frictional force: The frictional force, which is responsible for stopping the box, can be expressed as: $F_{\text{friction}} = \mu_k \cdot m \cdot g$ where $g = 9.81 \, \text{m/s}^2$.

2. Calculate acceleration: The frictional force equals mass times acceleration: $\mu_k \cdot m \cdot g = m \cdot a$ By canceling $m$ on both sides (since it's non-zero), we find: $a = \mu_k \cdot g = 0.45 \cdot 9.81 \approx 4.41 \, \text{m/s}^2$

3. Use kinematic equation: Since the box comes to rest, we have: $0 = v_i - a \cdot t$ Rearranging gives: $v_i = a \cdot t$ Plugging in known values gives: $v_i = 4.41 \, \text{m/s}^2 \cdot 5.0 \, \text{s} \approx 22.05 \, \text{m/s}$

4. Verify and Summarize

Thus, we can confirm that all calculations are correct and the relationships between forces and motion are appropriately applied. The initial velocity calculated is consistent with the given conditions.

Final Answer

The box’s initial velocity was approximately $22.05 \, \text{m/s}$.