Question

What is the coefficient of kinetic friction of a 75 kg box that slides down a 25° ramp with an acceleration of 3.6 m/s^2?

Answer 1

To determine the coefficient of kinetic friction for the box sliding down the ramp, we can start by analyzing the forces acting on the box.

The force of gravity acting on the box can be decomposed into two components: the force parallel to the ramp and the force perpendicular to the ramp. The force parallel to the ramp, also known as the component of the weight that contributes to the acceleration, is given by:

$F_{\text{parallel}}=m\cdot g\cdot \sin (\theta )$

where:

• $m$ is the mass of the box (75 kg),
• $g$ is the acceleration due to gravity (approximately 9.8 m/s^2), and
• $\theta$ is the angle of the ramp (25°).

Next, we can calculate the net force acting on the box along the ramp using Newton's second law:

$F_{\text{net}}=m\cdot a$

where:

• $a$ is the acceleration of the box (3.6 m/s^2).

Since the box is sliding down the ramp, the kinetic friction force ($F_{\text{friction}}$) opposes the motion and contributes to the net force. The kinetic friction force can be calculated using:

$F_{\text{friction}}=\mu \cdot N$

where:

• $\mu$ is the coefficient of kinetic friction (what we want to find), and
• $N$ is the normal force.

The normal force is equal to the force perpendicular to the ramp, which can be calculated as:

N=m⋅g⋅cos⁡(θ)N = m cdot g cdot cos( heta)N=m⋅g<span class="mspace" s