Solving Normal Force Problem: Find Fn with Ks=46.0 J, v=4.50m/s

[phy5ic5]

Homework Statement

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?

Homework Equations

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

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?

 

[phy5ic5]

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

 

[mtayab1994]

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

 

[phy5ic5]

Sorry about that- it's up there now.

 

[mtayab1994]

Sorry about that- it's up there now.

Is there any other drawing or no?

 

[phy5ic5]

The only figure given was the graph.

 

[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 : $$K(J)=46-23x$$

 

[phy5ic5]

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

 

[mtayab1994]

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

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

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

 

[phy5ic5]

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

 

[mtayab1994]

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

Were you given the answer?

 

[phy5ic5]

i wasn't given any answer

 

[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 :) : $$K_{s}=mgABcos(\frac{\pi}{2}+\theta)$$

 

[mtayab1994]

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

 

[phy5ic5]

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

 

[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?

 

[PeterO]

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

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.

 

[mtayab1994]

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.

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