A 2.61 kg block slides UP a 15.0° incline with k = 0.220. What is the friction force on the block?

Answer 1

Hello there! I assume that k=coefficient of kinetic friction ( I'm used to " u " ( pronouced " meuw " ) being used for the coefficient value. ) I could go into much more detail with the capability to draw a picture, but I'm new to the site. Nonetheless, I'll attempt to answer the question.

The force of kinetic friction ( Fk )=uN, where " N " represents the " normal force " of the block. The normal force of the block is the component of the blocks weight that is perpendicular to the surface upon which it slides.

At this point, I encourage you to draw a picture of the box on the incline. Next, draw a line from the center of the box that goes straight down to the ground ( perpendicular to the ground's surface ). This vector is the weight ( F=mg ) vector of the block.

Next, from the same line origin, draw a line that is perpendicular to the surface the box is sliding upon. This line represents the normal force, and it is at an angle to the vector representing the weight ( mg ) of the box, and that angle will be assigned a value of 15 degrees.

Returning to our original equation for the force of kinetic friction, Fk=uN. We are given the value of the coefficient of kinetic friction of u=0.220. We must now obtain a value for the normal force ( N ). Once we have the value of N, we can substitute N into the above equation to obtain the force of kinetic friction ( Fk ).

Now, draw a line from the same line origin as before, but allow this line to be parallel to the surface the block slides upon. NOW, USE YOUR IMAGINATION TO DRAG THIS LINE DOWN FAR ENOUGH BELOW THE BOX TO CREATE A RIGHT TRIANGLE. This vector should be perfectly perpendicular to the vector that represents the normal force ( N ). This line should also intersect the weight vector ( F=mg ) that was drawn.

You now have vectors that form a right triangle. The first line drawn ( straight downward )=the weight of the block ( F=mg ). This is the " hypotenuse " of your triangle. The vector parallel to the surface upon which the block slides is your " opposite " side. Additionally, the vector representing the normal force ( N ) is the " adjacent " side of your triangle.

Recall that the sin ( theta ) of a right triangle = ( opposite side ) / ( hypotenuse ), where " theta " is the angle of 15.0 degrees. The cos ( theta ) = ( adjacent side ) / ( hypotenuse ). Since the adjacent side of our triangle is represented by the N vector, we can use the cos ( theta ) function to solve for N. After we solve for N, we can plug our value into Fk=uN to solve the problem.

In this problem, the cos ( theta ) = ( N ) / ( mg ). Using algebra, we see that ( mg )( cos theta )=N. Since theta=15.0 degrees, N=( mg )( cos 15 degrees ). Therefore, if Fk=uN, and u=( 0.220 ), then Fk = ( 0.220 )( mg )( cos 15 degrees ).

ALWAYS REMEMBER TO CONVERT TO UNITS THAT ARE APPROPRIATE. We are told that the box has a mass of 2.61kg, but if we had been given a value in grams ( g ), we would have needed to convert grams to kilograms before solving the problem ( watch out for that on tests! ).

Now, Fk = ( 0.220 )( 2.61 kg )( 9.8 m/s^2 )( cos 15 degrees ). Unless some embarrassing and clumsy mistake was made somewhere, Fk = 5.45 N.