# Coefficient of kinetic energy

1. Oct 28, 2013

### ughschool

A box slides down a 30° ramp with an acceleration of 1.20 m/s^2. Determine the coefficient of kinetic friction between the box and the ramp

ok so the only equations I know are μk= Fk(kinetic friction)/Fn(normal) and that Fn=mass x gravity
so i dont know how to get Fk or mass... so help on solving this please?

im not really sure how to start other than maybe using the angle to find the x and y components of the ramp but again I dont know what I would do with that information
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 28, 2013

### rock.freak667

Well using the angle, what are the forces acting parallel to the ramp?

Then if you add up those forces they will give you a resultant force, what does Newton's 2nd Law say about these forces?

3. Oct 28, 2013

### ughschool

forces acting parallel to the ramp: gravity and friction?????

4. Oct 28, 2013

### rock.freak667

No, the force due to gravity acts vertically downwards, you can split this force into 2 components, one perpendicular to the ramp and another parallel to the ramp. Can you find these two forces? (given that the force due to gravity = weight = mg)

Also yes the force due to friction is parallel to the ramp.

5. Oct 28, 2013

### ughschool

Newton's 2nd law: Fnet=acceleration x mass

6. Oct 28, 2013

### ughschool

how would I find the force due to gravity if I dont have mass????

7. Oct 28, 2013

### rock.freak667

That's correct, so that is one part down, you just need to figure out the forces now in terms of mass and acceleration due to gravity.

For now, don't put in any numbers, just leave mass as 'm' and acceleration due to gravity as 'g' and angle as 'θ'.

8. Oct 28, 2013

### ughschool

ok so
Fnet= a x m
g in the x direction= m X 9.81 cosθ
g in the y direction= m x 9.81 sinθ
is that right???????

9. Oct 28, 2013

### ughschool

wait do i switch those? because my teacher has us rotate our axis so that the ramp is on the X axis

10. Oct 28, 2013

### rock.freak667

If you mean x-direction as parallel to the ramp and y as perpendicular, then yes.

How would you find the force due to friction using your equation with μk?

(Hint: the normal force = force perpendicular to the ramp)

11. Oct 28, 2013

### ughschool

ok so
Fn= m x 9.81 sinθ
so uk=Fk/m x 9.81 sinθ

im still not sure how that gives me Fk

12. Oct 28, 2013

### rock.freak667

Write Fk as the subject.

Now consider your forces in the x-direction i.e. parallel to the ramp. In what direction are the forces acting? (Which one will point down the ramp and which one up the ramp?)

After that, the question says that the box slides down the ramp, so is Fnet pointing down the ramp or up the ramp?

13. Oct 28, 2013

### ughschool

Fk=uk x (m x 9.81 sinθ)

left?

Fnet is pointing down the ramp?

14. Oct 28, 2013

### rock.freak667

Good

You have two forces parallel to the ramp, component of the weight and the friction, which one acts down the ramp and which up? When you figure this one, what would be the resultant of these two forces? (Use symbols as in A+B or A-B and so on. Not numbers)

This is correct.

15. Oct 28, 2013

### ughschool

weight is down the ramp and friction up right?

and the resultant is sqrt(a^2+b^2)?????

16. Oct 28, 2013

### rock.freak667

Yes, so if you had two forces pulling in opposite directions and you know the direction in which the resultant is in, what would you do to the two forces to get that resultant?

No.

17. Oct 28, 2013

### ughschool

18. Oct 28, 2013

### rock.freak667

Right but if one is opposing the other would it be positive or negative?

Using the forces you found above, add these up and what do you get?

19. Oct 28, 2013

### ughschool

a negative resultant

20. Oct 28, 2013

### rock.freak667

In terms of the forces, one will be positive and the other negative. Using the forces you found about, put one as negative and then add them up.

What do you get?

Also remember that this would be the resultant force which is the same as mass x acceleration or 'ma'.