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Question

https://phet.colorado.edu/sims/html/forces-and-motion-basics/latest/forces-and-motion-basics_en.html

What is the friction force on the box when a 50 N force is applied?

f = N

11. What is the maximum static friction force?

f = N

12. What is the normal force of the ground acting upward on the box?

N = N

13. Using the forces from the last two questions, find the coefficient of static friction. Enter at least two decimal places.

𝜇 =

14. What is the kinetic friction force?

f = N

15. Calculate the coefficient of kinetic friction. Enter at least 2 decimal places.

𝜇 =

Answer 1

a) the box does not accelerate when a 50N force is applied. Because there's no acceleration, there are no net forces, meaning that the force of friction must also be 50N.

b) check the sum of forces box in the top right. The maximum static friction is the same as the largest possible force that can be applied to the box while the net force is still 0, and the box is stationary. In this case, it's 125N.

c) The mass of the box is 50kg. The normal force will be mg, which is 50kg * 9.81m/s^2 which is 490.5N.

d) The coefficient of static friction is the maximum static frictional force divided by the normal force. 125N / 490.5N = .255

e) The force of kinetic friction is the force of friction acting on the box once it starts moving. You can find this by getting the box to move, and then decreasing the applied force until the net force = 0. At this point, the applied force is the same as the kinetic friction, which is 94N.

f) The coefficient of kinetic friction is the kinetic friction divided by the normal force. 94N/490.5N = .192

Hope this helped!