A stubborn, 191.0kg mule sits down and refuses to move. To drag the mule to the barn, the exasperated farmer ties a rope around the mule and pulls with his maximum force of 854.0N. The coefficients of friction between the mule and the ground are s=0.790 and k=0.280. What is the net force on the mule?

to start: net force=pulling-friction

= 854-191*g*.79

answer is 0

To find the net force on the mule, we need to consider both the applied force by the farmer and the forces of friction acting on the mule.

First, let's consider the static friction. The formula for static friction is given by:

F_static = μ_s * N

Where F_static is the static friction, μ_s is the coefficient of static friction, and N is the normal force.

In this case, the normal force acting on the mule is equal to its weight, which can be calculated as:

N = m * g

Where m is the mass of the mule and g is the acceleration due to gravity.

Substituting the given values, we have:

N = 191.0 kg * 9.8 m/s^2 = 1871.8 N

Now we can calculate the static friction:

F_static = 0.790 * 1871.8 N = 1477.40 N

Since the farmer is exerting a force of 854.0 N, which is less than the static friction force, the mule remains stationary. Therefore, the net force on the mule is zero.

It is important to note that if the applied force by the farmer exceeds the static friction force, the mule will start moving, and then we would need to consider kinetic friction. However, in this scenario, the mule remains stationary.