Finding final velocity using energy conservation with initial height and speed

[riseofphoenix]

Ok before we get to number 3, I need to show you what I did for number 2 because this is kind of a two-part problem:

PE_initial + KE_initial = [STRIKE]PE_final[/STRIKE] + KE_final
[STRIKE]m[/STRIKE]gy + (1/2)[STRIKE]m[/STRIKE]v_i² = (1/2)[STRIKE]m[/STRIKE]v_f²
gy + (1/2)v_i² = (1/2)v_f²
(9.81)(10) + (1/2)(10²) = v_f²/2
98.1 + 50 = v_f²/2
148.1 = v_f²/2
2(148.1) = v_f²
296.2 = v_f²
√(296.2) = v_f
17.2 m/s = v_f
(ANSWER "A")

So for this one, I didn't really know what to do :(
I tried doing this, but didn't get any of the choices above

PE_initial + KE_initial = [STRIKE]PE_final[/STRIKE] + KE_final
[STRIKE]m[/STRIKE]gy + (1/2)[STRIKE]m[/STRIKE]v_i² = (1/2)[STRIKE]m[/STRIKE]v_f²
gx + (1/2)v_i² = (1/2)v_f²
(9.81)(10) + (1/2)(25²) = (1/2)v_f²
98.1 + 312.5 = v_f²/2
410.6 = v_f²/2
2(410.6) = v_f²
821.2 = v_f²
√(821.2) = v_f
28.6 m/s v_f

But that's not right :(
Help?

 

[riseofphoenix]

Oh wait!

1) If μ = 1.5, then f_k = (1.5)(10 * -9.81)
f_k = -147.15

2) If f_k = -147.15, then F (forward) = +147.15

3) If F (forward) = +147.15, I can find acceleration, knowing that my mass, m, is 10 kg.

4) 147.15 = (10)a
147.15/10 = a
14.7 m/s² = a


5) Then I tried plugging into a kinematic equation that had the 3 variables I had the values for (v_f = 25 m/s, a = 14.7 m/s², and Δx = 9.0 m)

v_f² = v_i² + 2a_xΔx
25² = v_i² + 2*14.7*9
625 = v_i² + 264.6)
625 - 264.6 = v_i²
√(360.4) = v_i
18.98 = v_i
19.0 m/s = v_i


Am I right? :D

 

[SammyS]

...
https://www.physicsforums.com/attachment.php?attachmentid=53839&stc=1&d=1355269619

So for this one, I didn't really know what to do :(
I tried doing this, but didn't get any of the choices above

PE_initial + KE_initial = [STRIKE]PE_final[/STRIKE] + KE_final
[STRIKE]m[/STRIKE]gy + (1/2)[STRIKE]m[/STRIKE]v_i² = (1/2)[STRIKE]m[/STRIKE]v_f²
gx + (1/2)v_i² = (1/2)v_f²
(9.81)(10) + (1/2)(25²) = (1/2)v_f²
98.1 + 312.5 = v_f²/2
410.6 = v_f²/2
2(410.6) = v_f²
821.2 = v_f²
√(821.2) = v_f
28.6 m/s v_f

But that's not right :(
Help?

Treat this as an entirely different problem in which the block is moving with a velocity of 25.0 m/s on the horizontal portion of the track, just before encountering the 9 meter long section having non-zero friction.

In your second post, you did calculate the friction force correctly.

Since you used energy concepts to solve problem #2, that seems like a good approach for this problem as well.

I suggest using the Work - Energy Theorem.​

By the way: The potential energy is constant in problem #3. --- Why ?

 

[riseofphoenix]

Treat this as an entirely different problem in which the block is moving with a velocity of 25.0 m/s on the horizontal portion of the track, just before encountering the 9 meter long section having non-zero friction.

In your second post, you did calculate the friction force correctly.

Since you used energy concepts to solve problem #2, that seems like a good approach for this problem as well.

I suggest using the Work - Energy Theorem.​

By the way: The potential energy is constant in problem #3. --- Why ?

oh well I said it was constant because of this diagram I found online.

 

[riseofphoenix]

I have one more question though...

This one I have no idea how to solve for it... :(
It's a continuation of the last 2. It's basically the last part of this question.

Edit:

Oh wait!

Given
k = 50.4 N/m
v_i = 10 m/s
m = 10 kg

KE_{initial/moving block about to collide with spring in equilibrium} = PE_spring
(1/2)mv² = (1/2)kx²
(1/2)(10)(10²) = (1/2)(54)x²
500 = 27x²
500/27 = x²
18.5 = x²
√(18.5) = x
4.3 m = x

Is that right? :D

 

[haruspex]

Looks good.

 

[riseofphoenix]

Thanks!

 

[SammyS]

..

5) Then I tried plugging into a kinematic equation that had the 3 variables I had the values for (v_f = 25 m/s, a = 14.7 m/s², and Δx = 9.0 m)

v_f² = v_i² + 2a_xΔx
25² = v_i² + 2*14.7*9
625 = v_i² + 264.6)
625 - 264.6 = v_i²
√(360.4) = v_i
18.98 = v_i
19.0 m/s = v_i


Am I right? :D

That's sort of right .

Why did you call that v_i ?

That's the speed after passing over the friction region.

 

[haruspex]

Why did you call that v_i ?

That's the speed after passing over the friction region.

Which is the starting point of part 3.