Final answer:
To find the coefficient of friction between Mr. Frey and the ground, we apply Newton's second law and use the given force, mass, and acceleration. We calculate the components of the force applied, the normal force, and the frictional force that resists the motion. The coefficient of friction is found to be approximately 0.77.
Step-by-step explanation:
To solve for the coefficient of friction between Mr. Frey and the ground, we need to use Newton's second law of motion, which is F = ma, where F is the net force, m is the mass, and a is the acceleration. We are given the force that she pulls with (484 N), the angle (29.5°), Mr. Frey's mass (90 kg), and the acceleration (0.85 m/s²).
First, calculate the components of the force:
- Horizontal component, Fx = 484 N × cos(29.5°)
- Vertical component, Fy = 484 N × sin(29.5°)
The net force in the horizontal direction is the horizontal component of the pulling force minus the force of friction. The force of friction (f k) is given by f k = μk × N, where N is the normal force and μk is the coefficient of kinetic friction. The normal force is equal to the weight of Mr. Frey minus the vertical component of the pulling force, N = mg - Fy.
By using the equation Fx - f k = ma, and substituting f k with μkN, and solving for μk, we can find the coefficient of kinetic friction.
Let's calculate the components:
- Fx = 484 N × cos(29.5°) = 484 N × 0.867 = 419.738 N
- Fy = 484 N × sin(29.5°) = 484 N × 0.492 = 238.048 N
- N = mg - Fy = (90 kg)(9.8 m/s²) - 238.048 N = 882 N - 238.048 N = 643.952 N
Now, solve for the frictional force which resists the motion:
- f k = ma + Fx = (90 kg)(0.85 m/s²) + 419.738 N = 76.5 N + 419.738 N = 496.238 N
Therefore, μk = f k/N = 496.238 N / 643.952 N = 0.77. This is the coefficient of friction.