Randy pushes a 30.0 kg box resting on a rough surface by exerting a force of 85.0 N. The box did not move. What is the coefficient of static friction between the box and the surface where it lies?
μ > 85.0 / (30.0 * 9.80)
Wb =M*g = 30 * 9.8 = 294 N. = Fn.
Fap-Fs = M*a, 85-Fs = M*0 = 0,
Fs = 85 N.
u = Fs/Fn = 85/294 = 0.289.
To find the coefficient of static friction between the box and the surface, we can use the formula:
Coefficient of static friction (μ) = (Force of static friction) / (Normal force)
In this case, the force of static friction is equal to the force applied by Randy, since the box did not move. Therefore, the force of static friction is 85.0 N.
The normal force is the force exerted by the surface on the box in the upward direction. It is equal to the weight of the box, which can be calculated using the formula:
Normal force = mass × gravitational acceleration
Given that the mass of the box is 30.0 kg and the gravitational acceleration is approximately 9.8 m/s², we can calculate the normal force:
Normal force = 30.0 kg × 9.8 m/s² = 294 N
Now, we can substitute the values into the formula to find the coefficient of static friction:
Coefficient of static friction (μ) = 85.0 N / 294 N = 0.2898
Therefore, the coefficient of static friction between the box and the surface is approximately 0.2898.