# Normal Force problem

1. Dec 15, 2011

### phy5ic5

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

A block is sent up a frictionless ramp along which an x axis extends upward. The figure below gives the kinetic energy of the block as a function of position x; the scale of the figure's vertical axis is set by Ks = 46.0 J. If the block's initial speed is 4.50 m/s, what is the normal force on the block?

2. Relevant equations

Fn=mg
K=(1/2)m(v^2)

3. The attempt at a solution

Solved K=(1/2)m(v^2) for m and got 4.54 kg.
Plugged m into Fn=mg.

I got 44.5 N but this is not correct. What am I doing wrong?

#### Attached Files:

• ###### graph.gif
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Last edited: Dec 15, 2011
2. Dec 15, 2011

### phy5ic5

file:///C:/Users/Alyssa/Desktop/graph.gif

3. Dec 15, 2011

### mtayab1994

re-post the file as an attachment so i can see it please.

4. Dec 15, 2011

### phy5ic5

Sorry about that- it's up there now.

5. Dec 15, 2011

### mtayab1994

Is there any other drawing or no?

6. Dec 15, 2011

### phy5ic5

The only figure given was the graph.

7. Dec 15, 2011

### mtayab1994

Well, if you want to start, you will first have to draw a ramp and place your block on it, then draw a y axis and an x axis .

And you can represent the function as : $$K(J)=46-23x$$

8. Dec 15, 2011

### phy5ic5

I've done all the initial steps I just don't know how to get the Fn.

9. Dec 15, 2011

### mtayab1994

Well remember that Fn=mg only if an object is at rest on a horizontal ramp. In your case

$$F_{n}=mgcos(\theta)$$

10. Dec 15, 2011

### phy5ic5

but what would theta be? no angle was given...

11. Dec 15, 2011

### mtayab1994

Were you given the answer?

Last edited: Dec 15, 2011
12. Dec 15, 2011

### phy5ic5

i wasn't given any answer

13. Dec 15, 2011

### mtayab1994

ok to find the angle do the following:

$$E_{k(1)}-E_{k(2)}=W_{p}$$

and at 2 meters the kinetic energy is zero so that means the action stopped at 2 meters. You will have $$E_{k(2)}=0$$

and even more help :) : $$K_{s}=mgABcos(\frac{\pi}{2}+\theta)$$

Last edited: Dec 15, 2011
14. Dec 15, 2011

### mtayab1994

$$\frac{K_{s}}{-mgAB}=cos(\frac{\pi}{2}+\theta)$$

15. Dec 15, 2011

### phy5ic5

I still don't quite understand what I have to plug in and where.. Like where are the A and B coming from?

16. Dec 16, 2011

### mtayab1994

AB is the distance from the beginning of the action until the end of the action. In your case its 2 meters because the kinetic energy at 2 meters on the graph is 0; Wp is the work. Do you understand now?

17. Dec 16, 2011

### PeterO

As the block moves up the ramp, the kinetic energy is converted to potential energy.

You can thus work out how high the block reaches.
To reach that height it had to travel 2m up the ramp.
You can use trig to find the angle.

18. Dec 16, 2011

### mtayab1994

Yea, but since he's doing physics work he should find the angle by using the kinetic energy theorem.