Solving Friction Problem: 3.15 kg Block & 87.4 N Force

[BMWPower06]

Homework Statement

A 3.15 kg block is pushed along the ceiling with a constant applied force of F = 87.4 N that acts at an angle of alpha = 58.9° with the horizontal, as seen in the figure below.
http://aycu26.webshots.com/image/18345/2003241655315981977_rs.jpg
The block accelerates to the right at 5.49 m/s2. Determine the coefficient of kinetic friction between block and ceiling.

Homework Equations

Ff=uN

The Attempt at a Solution

Tried using the angle but don't know what to do.
Any ideas?

 

[hage567]

So what did you use for N? You need to show more of what you did. Have you drawn a diagram with all of the forces on it? Remember Newton's second law.

 

[m2003]

I would first assess the given information and identify the key variables and equations that can help solve the problem. From the given information, we can see that the block is being pushed with a constant force of 87.4 N at an angle of 58.9° with the horizontal. The block also has a mass of 3.15 kg and is accelerating at a rate of 5.49 m/s2.

To solve for the coefficient of kinetic friction, we can use the formula Ff = uN, where Ff is the force of friction, u is the coefficient of friction, and N is the normal force. From the figure, we can see that the normal force is acting perpendicular to the surface of the ceiling, so we can use the equation N = mg, where m is the mass of the block and g is the acceleration due to gravity (9.8 m/s2).

Next, we need to find the force of friction, which can be calculated using Newton's Second Law, F = ma, where F is the net force acting on the block and a is the acceleration of the block. In this case, the net force is the force of friction, so we can rewrite the equation as Ff = ma.

Putting all of this together, we get Ff = uN = umg = ma. We can rearrange this equation to solve for the coefficient of friction, u, by dividing both sides by mg. This gives us u = a/g.

Now, we can plug in the given values to solve for the coefficient of friction:

u = a/g = (5.49 m/s2)/(9.8 m/s2) = 0.56

Therefore, the coefficient of kinetic friction between the block and ceiling is 0.56. This means that the force of friction is 0.56 times the normal force, and it is the force that is opposing the motion of the block.