What is the relationship between speed and friction in physics?

[alicia113]

Question:

answers so far ( i don't know if its right that's why I am posting on here)

please let me know if I am doing it correctly thanks !

 

[alicia113]

also i have no idea how to do c! at all !

 

[PhanthomJay]

I don't know where your 'four meters' is coming from. $KE_i + PE_i = KE_f + PE_f$ , or $\Delta KE + \Delta PE = 0$ if you like. Please explain.

For part c, I don't see the rest of the pic after B. If it starts at 2.5 m, though, above the reference plane, where must it end up before it stops, without friction, using the same equation?
.

 

[alicia113]

Don't you add 1.5 and 2.5 together ?

 

[alicia113]

I don't know where your 'four meters' is coming from. $KE_i + PE_i = KE_f + PE_f$ , or $\Delta KE + \Delta PE = 0$ if you like. Please explain.

For part c, I don't see the rest of the pic after B. If it starts at 2.5 m, though, above the reference plane, where must it end up before it stops, without friction, using the same equation?
.

I have not attempted c and need to redo the beginning part. So what do I out as my delta h value then? Is everything else in that equation correct?

 

[PhanthomJay]

You are misreading the graph, and setting up your equation incorrectly as well.
The skater starts off with an initial speed of 2.6 m/s^2 at a height 2 .5 m above a reference plane. For part a, at point A, the skater is 1.5 m above ground at some speed. The delta h is -1.0 m. But be consistent in setting up the energy equation. Either use initial KE plus initial PE = final KE plus final PE, where the initial PE is 2.5mg and the final PE is 1.5mg, or use the "delta" formula I referenced, where the change in PE is (1.5mg - 2.5 mg) = -1.0mg.

 

[alicia113]

You are misreading the graph, and setting up your equation incorrectly as well.
The skater starts off with an initial speed of 2.6 m/s^2 at a height 2 .5 m above a reference plane. For part a, at point A, the skater is 1.5 m above ground at some speed. The delta h is -1.0 m. But be consistent in setting up the energy equation. Either use initial KE plus initial PE = final KE plus final PE, where the initial PE is 2.5mg and the final PE is 1.5mg, or use the "delta" formula I referenced, where the change in PE is (1.5mg - 2.5 mg) = -1.0mg.

The way i did it above is how my textbook told me to do it. I can take a picture and post it of you would liek. Of an example. Because I am very lost now. So I can use hate equation I just did above ?

 

[PhanthomJay]

The way i did it above is how my textbook told me to do it. I can take a picture and post it of you would liek. Of an example. Because I am very lost now. So I can use hate equation I just did above ?

Please use the first equation I gave you: 1/2m(2.6)^2 + mg(2.5) = 1/2mv^2 + mg(1.5). Solve for v at point A. Continue... Don't confuse h with delta h

 

[alicia113]

1/2(74.5)(2.6)^2 + (74.5)(9.8m\s)(2.5) = 1/2(74.5)v^2 + (74.5)(9.8m\s)(1.5)


so i solve for v^2 like i normally would with the above equation ?

 

[PhanthomJay]

Yes..

 

[alicia113]

and do the same for b... but replace with 1.5?

 

[PhanthomJay]

and do the same for b... but replace with 1.5?

No, start at the top again for initial condition (v = 2.6 and h_top = 2.5) and the final condition is at B (v = ? and h_B =0), which you sort of indicated the first time except you called it delta h instead of h_B, h_B is the height at B above the reference plane and is = to 0.

 

[alicia113]

So the value will be 0 instead of 1.5?0

 

[PhanthomJay]

Yes...1/2(74.5)(2.6)^2 + (74.5)(9.8)(2.5) = 1/2(74.5)v^2 + (74.5)(9.8)(0)

Solve for v.