Question

A stationary 17.5 kg object is located on a table near the surface of the earth. The coefficient of static friction between the surfaces is 0.42 and the coefficient of kinetic friction is 0.27. 1. A horizontal force of 40 N is applied to the object. a. What is the force of static friction?

Answer 1

The force of static friction can be calculated using the equation fs = μsN, where fs is the force of static friction, μs is the coefficient of static friction, and N is the normal force. In this case, the normal force is equal to the weight of the object, which is given by N = mg, where m is the mass of the object and g is the acceleration due to gravity. Therefore, the force of static friction is fs = μsmg. Substituting the given values into the equation, fs = (0.42)(17.5 kg)(9.8 m/s²) = 71.19 N.

Step-by-step explanation:

The force of static friction can be calculated using the equation fs = μsN, where fs is the force of static friction, μs is the coefficient of static friction, and N is the normal force. In this case, the normal force is equal to the weight of the object, which is given by N = mg, where m is the mass of the object and g is the acceleration due to gravity. Therefore, the force of static friction is fs = μsmg. Substituting the given values into the equation, fs = (0.42)(17.5 kg)(9.8 m/s²) = 71.19 N.