Relating position and force to speed

[Capncanada]

Homework Statement

The force graphed in figure is applied to a 1.8-{\rm kg} box initially at rest at x = 0 on a frictionless, horizontal surface.
(Graph is attached)

Homework Equations

F=ma (?)

The Attempt at a Solution

Don't know how to relate speed to these two scalars..

 

[cepheid]

You haven't stated what the problem is. At all.

 

[Capncanada]

I need to find the speed when position = 5, 10 and 15cm. Sorry.

 

[SammyS]

What level of math goes with your course?

 

[Capncanada]

It's physics 2A, I'm not having trouble with math I'm having trouble with finding the units of m/s out of units like force and position.

 

[SammyS]

Well, if it's calculus based, we might give one answer.

Trig. & algebra another.

Elementary algebra. ?

How is work done related to kinetic energy.

 

[Capncanada]

I had to take calculus to get into this course.

 

[cepheid]

A simple application of the work-energy theorem should get you on your way.

 

[SammyS]

What does area under the graph represent?

 

[Capncanada]

So F=ma can give me the acceleration of the object, I have no units of time to convert to speed here

 

[Capncanada]

The area represents work done?

 

[SammyS]

think about force time distance. Look at cepheid's suggestion.

 

[cepheid]

The area represents work done?

Yes.

 

[Capncanada]

I can solve for work done here but that leaves me with units of (kg*m^2)/s^2, I do not know how to get to speed here. This is probably very obvious, but I'm not seeing it.

 

[SammyS]

A simple application of the work-energy theorem should get you on your way.

Cap'ncanada,

What's this theorem?

 

[Capncanada]

F=ma?

 

[SammyS]

F=ma?

That's force.

Look at the units again. remember, you're looking for speed.

 

[Capncanada]

Err, W=KE=1/2mv^2?

Im having a lot of trouble with physics...

 

[cepheid]

F=ma?

No, that is not the work-energy theorem. That is Newton's 2nd Law. The work-energy theorem relates the work done on an object to its change in energy. If you haven't encountered it before (which seems strange) you can always look it up.

By the way, the work, of course, has dimensions of force*distance, therefore its units are Newton-metres = joules.

 

[cepheid]

Err, W=KE=1/2mv^2?

Im having a lot of trouble with physics...

That's pretty much it, but it should be that the work done on the object is equal to its CHANGE in kinetic energy. Therefore W = ΔKE = (1/2)mv_f² - (1/2)mv_i², where v_f and v_i are the final and initial velocities respectively.

 

[Capncanada]

but v_i will equal zero here yes?

I tried solving for when d= 5cm and it comes out and v_f comes out to 1m/s which is incorrect.

 

[cepheid]

but v_i will equal zero here yes?

I tried solving for when d= 5cm and it comes out and v_f comes out to 1m/s which is incorrect.

Okay. Well you understand all of the physics now, so it must just be an issue with arithmetic.
Can you post your solution for the x = 5 cm case? We can try and see where you went wrong.

Yes, v_i = 0 in this problem. Sorry if it seemed like I was being pedantic before. I was just making sure to get across the correct statement of the work-energy theorem.

 

[Capncanada]

No I appreciate the thoroughness of your explanations, I've been struggling all semester with this stuff and I need to understand it to a better extent, payed tutoring might be in my future. :(

Anyways,

W=F*d
W=18N*.05m
W=0.9J

W=KE=(1/2)mv^2
v^2=2KE/m
v^2=2(0.9J)/1.8kg
v=1m/s

 

[cepheid]

No I appreciate the thoroughness of your explanations, I've been struggling all semester with this stuff and I need to understand it to a better extent, payed tutoring might be in my future. :(

Anyways,

W=F*d
W=18N*.05m
W=0.9J

W=KE=(1/2)mv^2
v^2=2KE/m
v^2=2(0.9J)/1.8kg
v=1m/s

Well, it looks to me like the area underneath the curve is a triangle with a base of 5 cm and a height of ~18 N. Is the area of that triangular region just equal to base*height? Hint: what shape of region actually has an area of 18 N * 5 cm? What portion of that region is made up by the triangle?

 

[Capncanada]

Ahh... So I was supposed to divide the work I found in half since A=1/2b*h. Getting right answers now.

 

[cepheid]

Ahh... So I was supposed to divide the work I found in half since A=1/2b*h. Getting right answers now.

Yeah exactly. Your mistake was that W = F*Δx is NOT true if the force is not constant. In this case the force is a function of position F = F(x). The more general formula for work is W = ∫F(x)dx. That's why work is equal to the area under the force-position curve. In this case, since F(x) was linear, that curve was a triangle, and we could just evaluate the area geometrically (no need to integrate).

You can see that the more general formula reduces to the simpler one when the force is not varying. If F(x) = F = constant, then W = ∫F(x)dx = F∫dx = FΔx. You can see that, in that simpler case, the force-position curve would be flat, and the area underneath would be a rectangle (and simply multiplying base*height would work).

So, the summary is that the more general integration or "area under the curve" method always works, whereas the simpler "force*distance" method only works if the curve is flat (constant force).