Question

If one 53.0 kg student is sliding down an icy slope (30o from horizontal) with a constant acceleration of 2.00 m/s2 then what is the coefficient of kinetic friction µ? (g=9.8 m/s2) a) 0.658 b) 0.342 c) 0.255 d) 0.573

Answer 1

The coefficient of kinetic friction (µ) is approximately 0.255, which corresponds to option c).

Identify the forces involved:

Weight (mg): Acting downwards along the slope (53.0 kg * 9.8 m/s^2)

Normal force (N): Perpendicular to the slope, pushing back against the weight

Force of Friction (μkN): Opposing the direction of motion, proportional to the normal force

Resolve the weight force into components:

Weight downslope (mg sinθ): (53.0 kg * 9.8 m/s^2) * sin 30° ≈ 26.5 N

Weight normal to the slope (mg cosθ): (53.0 kg * 9.8 m/s^2) * cos 30° ≈ 45.6 N

Apply Newton's Second Law:

Force downslope - Friction = mass * acceleration

26.5 N - μ * 45.6 N = 53.0 kg * 2.00 m/s^2

Solve for μ:

Isolate μ: μ = (26.5 N - 53.0 kg * 2.00 m/s^2) / 45.6 N

μ ≈ 0.255

Answer 2

The coefficient of kinetic friction, µ, given that the 53.0 Kg student slid with a constant acceleration of 2 m/s², is 0.342 (option B)

How to calculate the coefficient of kinetic friction?

First, we shall calculate the frictional force acting on the student. This is shown below:

• Mass of student (m) = 53.0 Kg
• Acceleration (a) = 2 m/s²
• Net force = ma = 53 × 2 = 106 N
• Acceleration due to gravity (g) = 9.8 m/s²
• Angle of icy slope (θ) = 30°
• Parallel force on student = mgSineθ = 53 × 9.8 × Sine 30 = 259.7 N
• Frictional force =?

Frictional force = Parallel force - net force

= 259.7 - 106

= 153.7 N

Finally, we shall calculate the coefficient of kinetic friction. This is shown below:

• Frictional force (F) = 153.7 N
• Mass of student (m) = 53.0 Kg
• Acceleration due to gravity (g) = 9.8 m/s²
• Angle of icy slope (θ) = 30°
• Normal reaction (N) = mgCosθ = 53 × 9.8 × Cos 30 = 449.81 N
• Coefficient of kinetic friction (µ) =?

µ = F / N

= 153.7 / 449.81

= 0.342

Thus, the correct answer is option B