I need help with physics homework. Please explain the process so I can understand. (See details for question).

Question

A box slides down a 30.0 degree ramp with an acceleration of 1.20 m/s/s. Determine the Coefficient of kinetic friction between the box and the ramp.

Answer 1

The net force is equal to mass times acceleration

m*g*sin(30)-m*g*cos(30)*µk=m*1.2
solve for µk

µk=(g*sin(30)-1.2)/(g*cos(30))

µk=0.436

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Answer 2

The box has three forces acting on it: the force of gravitational attraction, magnitude mg, pulling the block toward the center of the earth; the force normal to the surface of the plane holding the block up; and the frictional force parallel to the surface of the plane.

For this particular problem, as is often the case for problems involving inclined planes, it helps to rotate the coordinate axes so that the x axis is parallel to the plane and the y axis is normal to it. It then becomes readily apparent that the gravitational force resolves into two components. The x component of magnitude, mg sin θ, tends to pull the block down the plane and is opposed by the kinetic frictional force component -kN, where k is the coefficient of kinetic friction and N is the normal force. The y component of the gravitational force, magnitude -mg cos θ, tends to press the block against the plane and is opposed by the normal force N.

Because there is no acceleration in the y direction (in the rotated coordinate system) N - mg cos θ = 0. Therefore, the magnitude of the normal force N is mg cos θ.

In the x direction, we have an unbalanced force, as is evidenced by the fact that the block is sliding down the plane at acceleration a, in this example, 1.20 m/s². Using Newton's second law, we find the force equation in this direction to be
mg sin θ - kN = ma
But we already know that N = mg cos θ, so we can make the substitution and get
mg sin θ - kmg cos θ = ma.

Solving for k, we get

k = tan θ - a/(g cos θ)
In this instance, θ is one of those special angles where the exact values of all the trig functions are easy to get:
tan θ = √3/3 and 1 / cos θ = 2√3/3.

Therefore,
k = √3/3 - a/g*2√3/3
Using 9.8 m/s for g, this gives
k = 0.44

Answer 3

The formula for the coefficient of kinetic friction is:

μk = Ff/Fn, where Ff is the frictional force and Fn is the normal force.

Consider the box which is sliding down the 30.0° ramp. The reason why it is sliding down is that there is a force that causes it to slide down. Now this force is equal to the component of the weight of the box which is parallel to the surface of the ramp and pointing downward.

Let's call that component Wx. Now Wx = Weight of Box x sin30.0°. (We tilt the x and y axes so that the x-axis is parallel to the ramp surface and the y-axis is perpendicular to the ramp surface.

The weight of the box is represented by a vector drawn vertically downward. Wy is the component drawn from the tail of the weight vector parallel to the y-axis, while Wx is the component drawn from the head of Wy (and forming right angle with it) parallel to the x-axis to the head of the weight vector. A right triangle is now formed, with the weight (W) as the hypotenuse and Wx and Wy as the legs. Now the angle opposite Wx is equal to 30.0°
Wx = (mg)sin30.0°
Wy = (mg)cos30.0°

Therefore, the forces acting along the x-axis (along the ramp surface) are: Wx (going down) and Ff(going up).

From Newton's Second Law of Motion:
Fnet = m a
Wx - Ff = ma
mgsin30.0° - μkFn = m(1.20 m/s^2)

Here, Fn = Wy = mgcos30.0°.
Then,
mgsin30.0° - μk(mgcos30.0°) = m(1.20 m/s^2)

Canceling m and solving for μk:

μk = [gsin30.0° - 1.20 m/s^2]/gcos30.0°
μk=[(9.8m/s^2)(sin30.0°-1.20 m/s^2]/(9.80 m/s^2)(cos30.0°)
μk = 0.44 ANS

Hope you understand.

teddy boy