# Magnitude of Acceleration and Work

1. Oct 10, 2012

### scotthyppo

Can someone help me
A 5kg box slides up an inclined plane surface, with an initial speed of 5.00 m/s. The plane makes a 30 degrees angle with horizontal. The coefficient of friction is .137. Assume g = 9.80 m/s^2.

A) what is the magnitude of the acceleration of the box?

B) What is the total work done by gravity on the box, from the start position, when the box has come to a stop?

2. Oct 10, 2012

### SHISHKABOB

what have you tried so far?

3. Oct 10, 2012

### scotthyppo

Ive tried finding the distance but could not

4. Oct 10, 2012

### SHISHKABOB

well so the first thing you'd want to do with a problem like this is set up a free-body diagram for the box

in that diagram you should break up the force due to gravity into its components and then also figure out the force due to friction

once you have the components of the force in the x-direction you can determine the net force in that direction and get the acceleration

5. Oct 10, 2012

### scotthyppo

Fy = 0
mg cos (30)

Fx =
ma + coefficient + mg sin (30)

What do you do after this?

6. Oct 10, 2012

### scotthyppo

For my X component I got a 55.3136

7. Oct 10, 2012

### SHISHKABOB

well so in the x direction you should have two forces: the force due to friction and the weight of the object that is parallel to the surface

the force due to friction is related to the normal force, and the normal force is equal to the weight of the object that is perpendicular to the surface

8. Oct 10, 2012

### scotthyppo

I am still confused can you show me what you mean?

9. Oct 10, 2012

### SHISHKABOB

do you have any examples from your notes where your teacher did an object on an incline?

10. Oct 10, 2012

### scotthyppo

11. Oct 10, 2012

### SHISHKABOB

he did an object on a downward slope, you mean?

12. Oct 10, 2012

### scotthyppo

Yep
This is what I did for my calculation:
Fy = N - mg cos (30) = 0
N = mg cos (30)
N = 42.4352
Fx I did:
(42.4352x .137) x (5x9.8) x sin(30)
and got 142.434
x direction you should have two forces: the force due to friction and the weight of the object that is parallel to the surface
How do you calculate that

13. Oct 10, 2012

### SHISHKABOB

well so where you calculate Fx it seems like you've got mgsin(theta) which is the weight of the object that is parallel to the surface

the frictional force is also there where you multiply the normal force by the coefficient of friction

but it seems like you multiplied the frictional force by the parallel weight; what you want to do is *add* them together. But of course remember that they are both negative, because they are pointing in the opposite direction of the velocity of the box.

14. Oct 10, 2012

### scotthyppo

So
-mg sin(30) - mew = max
-mg sin (30) - .137 mg cos (30) = max
So the m cancels on both side
-g sin (30) - .137 g cos (30) = ax
acceleration = -6.063

For the total work done by gravity on the box
W = F (cos (30))
V^2 = Vo^2 + 2ax
0 = 5^2 + 2 (-6.063)x
= 25 + (-12.126) x
X= 2.06
W = mg (cos 30) 2.06
W= 5 (9.8) (cos 30) 2.06
W = 87.48813 J

Is this right?

15. Oct 10, 2012

### SHISHKABOB

yes it looks good

but just remember that in part a it asks for the *magnitude* of the acceleration

16. Oct 10, 2012

### scotthyppo

What do you mean by that?
So the magnitude of the acceleration should just be 6.063

17. Oct 10, 2012

### SHISHKABOB

remember that a vector has both a magnitude and a direction. In this case, there are only two directions: positive x direction (up the slope) and negative x direction (down the slope).

Since the acceleration is a vector, it has both a magnitude and a direction. In this case, the direction of the acceleration is always *down the slope*. So its direction is negative, but the magnitude of a vector is always just some number.