Kinetic Frintion

## Question:

I've tried over and over to get this, but I'm missing something. Can you offer any assistance? A 2kg block is released from rest on a 45 degree incline and slides a distance of 1.5m down the incline in 2s. What is the coefficient of kinetic friction between the block and the incline?

By the way, I'm having difficulties here and there in physics. Are there any webpages or books with lots of examples that you recommend that will help me?

Thanks

You might try TIPTOP for additional help.

Regarding your specific problem, we will apply a trick that will help you with any problem. Begin with the definition of the quantity you are asked to find. The coefficient of kinetic friction is defined as the ratio of the force of friction divided by the force holding the sliding surfaces together. Now we go after the components of the definition.

First the force holding the surfaces together, called the normal force. If the plane were at an angle of zero degrees, the normal force would be simply the weight of the object. If the plane were vertical the normal force would be zero. With the plane at some angle between zero and ninety degrees the normal force will depend on the angle. Angular dependence implies trigonometry so we need to multiply the weight by a trigonometric function which is 1.0 at 0.0 degrees and 0.0 and 90.0 degrees. The cosine meets our specifications so the normal force is the weight times the cosine of the plane's angle (0.707 for 45 degrees). The weight is the mass (2.0kg) times the acceleration due to gravity (9.8m/s/s). This gives a normal force of 2.0*9.8*0.707=13.857 Newtons.

Next we need the force of friction. Here we need a definition of the force of friction in terms of the information that is available. We are looking for a force and we are given a distance and a time. We know a relationship between acceleration, distance and time so we could get an acceleration. Acceleration and force are related by mass so knowing the mass we can find the net force acting on the block. The total force on the block is made up of the force of the Earth on the block and the force of the plane on the block. In the direction perpendicular to the plane these forces exactly cancel out. In the direction along the plane they are the force of friction which we are looking for and the force of gravity along the plane.

The force of gravity on the block, along the plane would be 0.0 at 0.0 degrees and the weight of the block at 90.0 degrees. The trig function that varies in this way is the sine function so the force along the plane is the mass of the block times the acceleration due to gravity times the sine of the plane's angle. In this case (45 degrees), the sine equals the cosine, 0.707, so the force due to gravity along the plane is 2.0*9.8*0.707=13.857 Newtons.

The acceleration (a) of the block is in the expression d=1/2*a*t2 where d is the distance moved (1.5m) and t is the time it took (2.0) seconds. From this we see that 1.5=1/2*a*4 or a=3m/s/s. The net force on the block then is its mass times the net acceleration or 3.0m/s/s*2.0kg=6 Newtons.

We know that the force of friction is always opposite the direction of motion so the net force is given by the force of gravity along the plane minus the force of friction. Here 6.0 Newtons=13.857 - Force of friction. Solving this for the force of friction gives Ff=7.857 Newtons.

Now we have the components of the definition of the kinetic friction coefficient. It is 7.587/13.587=0.558.

I have included a lot of explanation and a general approach to solving problems. I hope this helps. Let me know if you do not understand.

JDJones M. Casco Associates