The 3.1 kg box shown below slides with a constant speed down an incline at an angle of θ = 28.0° to the horizontal.
What is the coefficient of kinetic friction?
Wb = M*g = 3.1 * 9.8 = 30.4 N.
Fp = 30.4*sin28 = 14.3 N. = Force parallel to the incline.
Fn = 30.4*Cos28 = 26.8 N. = Normal force
Fp-Fk = M*a.
Fp-Fk = M*0 = 0.
Fk = Fp = 14.3 N. = Force of kinetic
friction.
u = Fk/Fn = 14.3/26.8 = 0.533.
To determine the coefficient of kinetic friction, we need to analyze the forces acting on the box sliding down the incline.
1. Start by drawing a free-body diagram of the box. Identify the forces:
- The weight force, mg, pointing vertically downwards.
- The normal force, N, perpendicular to the incline.
- The force of kinetic friction, fk, parallel to the incline.
- The force component, Fx, along the x-axis of the incline.
- The force component, Fy, along the y-axis perpendicular to the incline.
2. Based on the information given, we know that the box is sliding down the incline at a constant speed. This implies that the net force acting on the box is zero. Therefore, we can write down the equation:
Fx = fk = μk * N, where μk is the coefficient of kinetic friction.
3. Calculate the gravitational force component along the incline, Fx. We can express this force using trigonometry:
Fx = mg * sin(θ), where θ is the angle of the incline.
4. Calculate the gravitational force component perpendicular to the incline, Fy:
Fy = mg * cos(θ), where θ is the angle of the incline.
5. Since the box is sliding at a constant speed, we know that the force of kinetic friction is equal in magnitude but opposite in direction to the force component along the x-axis:
fk = Fx = mg * sin(θ).
6. Finally, substitute the value of fk into the equation from step 2 to find the coefficient of kinetic friction:
μk = fk / N = (mg * sin(θ)) / (mg * cos(θ)) = tan(θ).
Therefore, the coefficient of kinetic friction is equal to the tangent of the angle of the incline, which in this case is tan(θ) = tan(28.0°).