Block on an incline with Friction

[love2dream89]

Homework Statement

A block with a mass of m = 2.29 kg is at rest on an incline. The angle of the incline is θ = 15.9° with respect to the horizontal. The coefficient of static friction between the object and the incline is μs = 0.579, the coefficient of kinetic friction is μk = 0.389.

What would be the magnitude of the object's acceleration after an initial impulse set it in motion uphill on the incline?

Homework Equations

Newtons Second Law
F= ma
??

The Attempt at a Solution

[-(coefficient k)mg-mgsin(theta)]/m and then making that answer positive because it is a magnitude. Where am I going wrong here?

 

[Dick]

You probably aren't actually drawing a force diagram. The frictional force is proportional to the normal force, which isn't mg. The normal force depends on the angle too.

 

[nlightner]

Your attempt at a solution is on the right track, but there are a few things to consider. First, the given coefficients of friction are for static and kinetic friction, which means that the block is either at rest or in motion. Therefore, you will need to consider both cases separately.

For the case of static friction, the maximum force of static friction that can act on the block is given by μsN, where N is the normal force exerted by the incline on the block. This normal force can be calculated using trigonometry as N = mgcos(theta). Therefore, the maximum force of static friction is μsmgcos(theta). Since the block is initially at rest, this force must be equal and opposite to the force pulling the block down the incline, which is mgsin(theta). Therefore, the maximum force of static friction can be written as μsmgcos(theta) = mgsin(theta), which can be rearranged to find the normal force N = mg/(cos(theta)+μssin(theta)).

For the case of kinetic friction, the force of kinetic friction is given by μkN, where N is again the normal force. Since the block is now in motion, the force of kinetic friction will act in the opposite direction of motion, which is up the incline. Therefore, the acceleration of the block can be calculated using Newton's second law as a = (μkN-mgsin(theta))/m. Plugging in the expression for N found above, we get a = (μkmg/(cos(theta)+μksin(theta))-gsin(theta))/m.

To find the magnitude of the acceleration, we can use the Pythagorean theorem to combine the x and y components of the acceleration (since the incline is at an angle). The magnitude of the acceleration can be written as a = √(ax^2+ay^2), where ax is the acceleration in the x-direction (parallel to the incline) and ay is the acceleration in the y-direction (perpendicular to the incline). Plugging in the expressions for ax and ay found above, we get a = √((μsmgcos(theta)-mgsin(theta))^2+(μkmg/(cos(theta)+μksin(theta))-gsin(theta))^2)/m.

Therefore, to find the magnitude of the acceleration, you will need to plug in the given values for m, θ, μ