One more physics question?

Question

lol everyone's answers have been so helpful when i asked about physics yesterday i thought id throw one more out there

this one is pretty simple, i think.. its just that neither my textbook OR the notes from class show an example similar to this, so i dont really know where to start...

"a child slides down a slide with a 28 degree incline, and at the bottom her speed is precisely half of what it would have been if the slide had been frictionless. Calculate the coefficient of kinetic friction between the slide and the child."
i might need to use

force of friction = coefficient of kinetic friction x normal force

but idk an equation that relates with angles!

Answer 1

I would tackle this one in terms of conservation of energy.

> "...at the bottom her speed is precisely half of what it would have been..."

This means some of her energy was lost due to friction. When there's no friction, you have this:

ΔPE = ΔKE
or:
mgh = mv²/2
or:
v_no_friction = sqrt(2gh)

But her actual speed was only half that:
v_friction = sqrt(2gh)/2

So her KE at the bottom is:
m(v_friction)²/2 = m(2gh/4)/2 = mgh/4

So, at the top of the slide, she had this much energy (PE):
mgh

But at the bottom she only had this much energy (KE):
mgh/4

That means she LOST this much due to friction:
mgh – mgh/4 = (3/4)mgh

Now, the formula for energy lost due to friction is:

energy loss = F_friction · d

Where F_friction is the force of friction, and d is the distance over which the friction acts (in this case, the length of the slide. So:

energy loss = F_friction · d = (3/4)mgh

Now make these substitutions:

F_friction = F_normal(μ) [where μ = coefficient of friction]
h = height of slide = (length)(sin28) = d(sin28)

So now we have:
F_normal(μ) · d = (3/4)mgd(sin28)

Divide both sides by "d", and "d" magically disappears:
F_normal(μ) = (3/4)mg(sin28)

Whenever something's sitting/sliding on an incline of θ degrees, F_normal = weight(cosθ) = mg(cosθ). So:

mg(cos28)(μ) = (3/4)mg(sin28)

Divide both sides by mg(cos28) and we get:

μ = (3/4)(tan28)

Answer 2

The process of physics is not the process of looking up equations in the textbook--- it's the process of *deriving* the equations on your own for each particular system. Every time you do a force problem, the process is the same, but the equations you derive will be different.

1. Draw a picture of the box on the incline. Then draw vectors representing all the forces that are acting on the box. Gravity, friction, etc.

2. Choose x and y axes. You want to try to choose axes so that the expected motion of the box is along the x-axis if you can.

3. Resolve all vectors into X and Y components.

4. Apply Newton's Second Law, F=ma, in each of the x,y, and z directions. Plug in forces that act in each direction, and zero for a when there is no acceleration along that axis.

5. You've just derived the equations of motion of the box. Solve the system for whatever unknown you need.

Every force-on-a-box problem is done this way. It may get complicated with more boxes, forces that must be equal to other forces, accels that are equal (or opposite) to other accels, but all that means is more equations to derive in the same way. Once you have your system of equations, it's all algebra from there.

Answer 3

x axis (with friction) (axis is parallel to incline):

ΣF = -Ff + Fg =ma
-μN + mgsin(28) = ma
-μ(mgcos(28)) + mgsin(28) = ma
μ = (a – gsin(28)) / -gcos(28)

x axis (without friction):

ΣF = Fg =ma
mgsin(28) = ma
gsin(28) = a
a = 4.61

If “v” without friction is 2 times “v” with friction, then “a” without friction is 4 times “a” with friction. (You can check this with the Equations of Motion: Vf^2 = Vi^2 +2ad)

So “a” with friction = 4.61/4 = 1.15

Substitute new acceleration into equation for “μ”

μ = (1.15 - 9.81(sin(28))/-9.81cos(28)
μ = 0.40

μ is a dimensionless number

It's hard to explain the angle equation without a picture. It can get real confusing real quick. It took me a lot of studying to figure it all out.

Answer 4

You have two direction horizontal and vertical.

Gravity is acting in a vertical direction.

The angle comes in as the ratio of one to the other. Heard of Sine, Cos, etc? You have to 'build' your 'own equation'.

If the slide is frictionless, what will impede the velocity imparted by gravity?

Answer 5

When there is no friction
mg h = 0.5mv1^2
When there is friction
mg h = 0.5mv2^2 + μ mg cos θ *L
Since v2 = v1 /2 , we get
mg h =[ 1/4 ] [0.5mv1^2] + μ mg cos θ *L
mg h =[ 1/4 ] mg h + μ mg cos θ *L
[3/4] h = μ cos θ *L
[3/4] L sin θ = μ L cos θ since h = L sin θ
μ = [3/4] tan θ
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