Work on Block in Vertical Circle: MasteringPhysics

[kottur]

Homework Statement

A small block with mass 0.0325 kg slides in a vertical circle of radius 0.475 m on the inside of a circular track. During one of the revolutions of the block, when the block is at the bottom of its path, point A, the magnitude of the normal force exerted on the block by the track has magnitude 3.80 N. In this same revolution, when the block reaches the top of its path, point B, the magnitude of the normal force exerted on the block has magnitude 0.680 N.

How much work was done on the block by friction during the motion of the block from point A to point B?

Homework Equations

K1+U1+W=K2+U2

W=Fs

Conserved energy and motion in a circle

The Attempt at a Solution

I know that the distance from A to B is 2*r*pi=(19/20)*pi and I can use that as s in W=Fs. I don't know how to find out the friction.
All help would be very appreciated!

 

[Doc Al]

I know that the distance from A to B is 2*r*pi=(19/20)*pi and I can use that as s in W=Fs. I don't know how to find out the friction.

That's the hard way. Try the easy way: Compare the mechanical energy at A and B.

 

[kottur]

Should I do it with change in velocity?
At the top: N+mg=m(v^2/r)=0.0325kg(v^2/0.475m)=0.680N so v=3.15 m/s.
At the bottom N-mg=m(v^2/r)=0.0325kg(v^2/0.457m)=3.80N so v=7.31 m/s.
I got it from here: http://dev.physicslab.org/Document.aspx?doctype=3&filename=OscillatoryMotion_VerticalCircles.xml

Is this in the right direction?

 

[Doc Al]

Should I do it with change in velocity?
At the top: N+mg=m(v^2/r)=0.0325kg(v^2/0.475m)=0.680N so v=3.15 m/s.
At the bottom N-mg=m(v^2/r)=0.0325kg(v^2/0.457m)=3.80N so v=7.31 m/s.
I got it from here: http://dev.physicslab.org/Document.aspx?doctype=3&filename=OscillatoryMotion_VerticalCircles.xml

Is this in the right direction?

The equations N ± mg = mv^2/r are correct, but your calculations are not.

But that's the right idea to find the speed. (Even better is to find the KE, which is what you really want.)

Again, you'll eventually want to compare the total mechanical energy at A and B to see how much is lost.

 

[kottur]

Ok I tried finding KE with Wtot=K2-K1=0.5mv2^2-0.5mv1^2=0.5*0.0325kg*(7.31m/s)^2 - 0.5*0.0325kg*(3.15m/s)^2=0.7071J which is not correct...

I don't understand what's wrong with my calculations from above because I just plugged in the given variables to the formula.

 

[Doc Al]

I don't understand what's wrong with my calculations from above because I just plugged in the given variables to the formula.

What formula did you use?

 

[kottur]

The equations N ± mg = mv^2/r are correct, but your calculations are not.

You said that my calculations were wrong. I don't understand why because I just put the given variables into the equation N ± mg = mv^2/r.

 

[Doc Al]

You said that my calculations were wrong. I don't understand why because I just put the given variables into the equation N ± mg = mv^2/r.

Well, not exactly. Let's see what you actually did:

At the top: N+mg=m(v^2/r)=0.0325kg(v^2/0.475m)=0.680N so v=3.15 m/s.

The first part is correct: N+mg=m(v^2/r)=0.0325kg(v^2/0.475m)

But this does not equal 0.68 N.

You essentially used this formula: N = m(v^2/r), when you should have used: N+mg=m(v^2/r).

You left off the mg.

 

[kottur]

Okay thank you.

Now I got: N+mg=m(v^2/r)=0.0325kg(v^2/0.475m)+(9,8m/s^2)*0.0325kg =0.680N+0.3185N=0.9985N so v=3.82 m/s.

Then I got: Wtot=K2-K1=0.5mv2^2-0.5mv1^2=0.5*0.0325kg*(3.82m/s)^2 - 0.5*0.0325kg*(7.31m/s)^2= -0.631J which is still not correct.

Should I be using a different formula?

 

[Doc Al]

Now I got: N+mg=m(v^2/r)=0.0325kg(v^2/0.475m)+(9,8m/s^2)*0.0325kg =0.680N+0.3185N=0.9985N so v=3.82 m/s.

This is good for point B. Now fix the other equation you used, for point A.

Then I got: Wtot=K2-K1=0.5mv2^2-0.5mv1^2=0.5*0.0325kg*(3.82m/s)^2 - 0.5*0.0325kg*(7.31m/s)^2= -0.631J which is still not correct.

Two problems:
- You're still using your incorrect value for the speed at point A.
- You're neglecting potential energy.

 

[kottur]

0.0325kg(v^2/0.475m)=3.80-mg=3.4815N so v=7.13m/s.

Wtot=K2-K1=0.5mv2^2-0.5mv1^2=0.5*0.0325kg*(3.82m/s)^2 - 0.5*0.0325kg*(7.13m/s)^2= -0.589J

Is this enough for the correct answer? Do I still need to include the potential energy or have I done that already?

I'm sorry about being slow, I'm just so confused...

 

[Doc Al]

0.0325kg(v^2/0.475m)=3.80-mg=3.4815N so v=7.13m/s.

Good.

Wtot=K2-K1=0.5mv2^2-0.5mv1^2=0.5*0.0325kg*(3.82m/s)^2 - 0.5*0.0325kg*(7.13m/s)^2= -0.589J

Is this enough for the correct answer? Do I still need to include the potential energy or have I done that already?

You still need to include gravitational potential energy. Call the potential energy at the bottom position = 0.

Use this formula, which you had in your first post: K1+U1+W=K2+U2

 

[kottur]

K1+U1+W=K2+U2
K1+U1+W=K2+mgh
0.5*0.0325kg*(7.13m/s)^2+0+W=0.5*0,0325*(3.82m/s)^2+0.0325kg*9.8m/s^2*0.95m
so W= -0.286J

Thank you ever so much for the help. :)