1)A logging truck is dragging a 427 kg log through the forest at a constant speed of 3.5 m/s. If the truck is applying a force of 1047 N to the log to keep it moving, what is the coefficient of friction between the log and the ground?
im thinking you use the kinetic friction= μk(normal force) equation so does that mean it would be 3.5 m/s=μ(1047N) but then where does the 427 kg come in??
The speed has nothing to do with it. You just need to know that it is constant. That means the net force is zero.
Towing force = Friction force
1047 N = M*g*μk
= 427*9.8*μk
Solve for μk
To find the coefficient of friction between the log and the ground, let's break down the problem step by step.
1) Start by drawing a free-body diagram for the log. The forces acting on the log are:
- The force of gravity (weight) acting downward, which is equal to the mass of the log multiplied by the acceleration due to gravity (Fg = m * g).
- The normal force acting upward, which is equal in magnitude but opposite in direction to the force of gravity. In this case, since the log is on a level surface and not accelerating vertically, the normal force is equal to the weight of the log (Fn = Fg).
2) Since the truck is applying a force of 1047 N to the log to keep it moving at a constant speed, we know that the frictional force (Ff) acting on the log is equal in magnitude but opposite in direction to the applied force (Ff = -1047 N).
3) The frictional force can be expressed as the product of the coefficient of friction (μ) and the normal force (Fn). So, Ff = μ * Fn.
4) Combine steps 2 and 3 to get the equation -1047 N = μ * 427 kg * g.
5) Rearrange the equation to isolate the coefficient of friction (μ). The acceleration due to gravity (g) is approximately 9.8 m/s^2. So, the equation becomes μ = -1047 N / (427 kg * 9.8 m/s^2).
6) Calculate the coefficient of friction using the given values: μ ≈ -0.255
Note: The negative sign indicates that the frictional force opposes the motion of the log, which is consistent with the log being dragged by the truck. The coefficient of friction is always positive, so the magnitude of the coefficient would be 0.255.