# I need help on a force question please

1. Apr 22, 2007

### skittlez411

i need help on a force question please!!

1. The problem statement, all variables and given/known data

The 1.0kg physics book in the figure below is connected by a string to a 500g coffee cup. The book is given a push up the slope and released with a speed of 3.0 m/s. The coefficients of friction are us=0.50 and uk=0.20.

a) How far does the book slide?
b) At the highest point, does the book stick to the slope or does it slide back down?

2. Relevant equations

F=mgsin(theta)
F=ukmg

im not too sure what equations to use cuz im kinda confused at the free body diagram.

3. The attempt at a solution

F=mgsin20
= (1.0)(9.8)(sin20)
= 3.35N

Do I have to split the 2 objects? like draw one FBD for the cup, and one for the book? How do i do that? Please help!

Last edited: Apr 22, 2007
2. Apr 22, 2007

### Mindscrape

What does Newton's 3rd law tell you?

3. Apr 22, 2007

### skittlez411

for every interaction, there's an equal or opposite reaction.

4. Apr 22, 2007

### Mindscrape

Right, so what would the action/reaction pair for the tension of the rope be?

5. Apr 22, 2007

### skittlez411

it balances??

6. Apr 23, 2007

### skittlez411

anybody that can help??

7. Apr 23, 2007

### mezarashi

What you should first do is draw a free-body diagram of the book. What are the forces acting on the book? In which direction is the net resultant force vector?

From here, you can use two approaches to find the distance traveled by the book. Either (1) find the acceleration of the book and use the initial velocity in the kinematics equation or (2) use the work-energy equations.

8. Apr 23, 2007

### skittlez411

yah i was thinking abt those 2 approaches too. but then i got stuck becuz i dont have the final velocity and the time. so how do i do that?

I drew the FBD for the book and found that the force acting on the book is F=(1.0)(9.8)(cos20)=9.2N

9. Apr 23, 2007

### denverdoc

well if you were to do w-e theorum,

you would start with the initial energy of the book and cup, in this case KE and equate it with the change in potential energy of the book and cup (careful here keeping track of displacements in vertical direction) when velocity is zero. Now that assumes no friction.

The frictional work would be the frictional force times the distance it moves up the ramp. Lets call this D. This D and the distance the cup is lifted are the same, but it is not quite the same D as you would use for the book in computing its change in potential energy. But the two are related.

If you were to calculate acceleration,
you would need to account for three forces acting on the book; you have mentioned one, but there are three, the Tension, the frictional force and the fraction of the weight that works to pull the book down the slope. It is not cosine 20 if i read the figure right. If it were 0 degrees, that is horizontal, the weight of the book would not come into play except via frictional force.