An 18.0 kg box is released on a 35.0° incline and accelerates down the incline at 0.269 m/s2. Find the friction force impeding its motion.
How large is the coefficient of friction?
To find the friction force impeding the motion of the box, we need to use the equation:
Ffriction = μ * N
where Ffriction is the friction force, μ is the coefficient of friction, and N is the normal force.
First, let's determine the normal force acting on the box. The normal force is the perpendicular force exerted by a surface to support the weight of the object resting on it. In this case, the normal force can be defined as:
N = mg * cosθ
where m is the mass of the box, g is the acceleration due to gravity (approximately 9.8 m/s^2), and θ is the angle of the incline (35.0°).
N = (18.0 kg) * (9.8 m/s^2) * cos(35.0°)
N ≈ 152.57 N
Next, we can substitute the value of the normal force into our friction force equation:
Ffriction = μ * N
Ffriction = μ * 152.57 N
We also know that the acceleration down the incline is 0.269 m/s^2. The net force acting on the box down the incline is the component of the gravitational force down the incline minus the friction force. So,
Fnet = m * sinθ * g - Ffriction
ma = mg * sinθ - μ * N
Now, we can rearrange the equation to solve for the coefficient of friction μ:
μ = (m * sinθ * g - ma) / N
μ = [(18.0 kg) * sin(35.0°) * (9.8 m/s^2) - (18.0 kg) * (0.269 m/s^2)] / 152.57 N
By substituting the values and evaluating the expression, we can find the coefficient of friction.