A student with a mass of 66kg is rollerblading at a velocity of 7.8m/s[W] when the trail suddenly becomes very rough and as a result the student stops pushing and comes to a stop in 0.95s.
Determine the net force and acceleration of the student on the rough trail.
Calculate the coefficient of kinetic friction between the rollerblades and the rough trail.
momentum = 66*7.8
average force = change in momentum/time
= -66*7.8/.95 Newtons
acceleration = average force/ 66
friction force = average force
= mu m g
so
mu = |acceleration|/g
To determine the net force and acceleration of the student on the rough trail, you can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration, F = m * a.
1. Calculate the net force:
Given:
Mass of the student, m = 66 kg
Velocity of the student, v = 7.8 m/s (assuming it becomes 0 m/s after stopping)
The initial velocity of the student is in the westward direction, and the final velocity is 0 m/s. The change in velocity (Δv) is the final velocity minus the initial velocity: Δv = 0 m/s - 7.8 m/s = -7.8 m/s.
Since the direction of the acceleration is opposite to the direction of motion, the acceleration (a) can be calculated as a = Δv / t, where t is the time taken to come to a stop:
Given:
Time taken to stop, t = 0.95 s
Substituting the values:
a = (-7.8 m/s) / (0.95 s) = -8.21 m/s^2 (negative sign indicates acceleration is in the opposite direction)
Using Newton's second law, F = m * a:
F = (66 kg) * (-8.21 m/s^2) = -541.86 N (negative sign indicates the force acts in the opposite direction)
Therefore, the net force acting on the student on the rough trail is approximately -541.86 N.
2. Calculate the coefficient of kinetic friction:
The force of kinetic friction can be calculated using the equation F_friction = μ_k * m * g, where μ_k is the coefficient of kinetic friction, m is the mass of the student, and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Given:
Mass of the student, m = 66 kg
Acceleration due to gravity, g = 9.8 m/s^2
Substituting the values:
F_friction = μ_k * (66 kg) * (9.8 m/s^2)
Since the net force acting on the student is equal to the force of friction, we can equate the two:
-541.86 N = μ_k * (66 kg) * (9.8 m/s^2)
Simplifying and solving for μ_k, we get:
μ_k = -541.86 N / (66 kg * 9.8 m/s^2)
μ_k ≈ -0.862
The coefficient of kinetic friction between the rollerblades and the rough trail is approximately -0.862. The negative sign indicates that the friction force is acting in the opposite direction to the motion of the student.