Forces and Tension- given coefficient of friction to find speed?

Oct 2015
20
0
Okay, I've posted a lot, but this is the lat problem I'm having an issue with. Here I go!

Screen Shot 2015-10-11 at 4.08.43 PM.png

i) I have absolutely no idea. I can't even attempt because I honestly don't know what to do or where to start. How the heck am I supposed to find speed with what's given? It's not like I have time or distance to help me here. I absolutely need help here.

ii) my numbers were:
coefficient of friction=0.1
mass 1 = 2 kg
mass 2 = 5 kg
g = -9.8 m/s^2
F normal = 5(9.8) = 49 N
kinetic friction = 49 N (0.1) = 4.9 N

now solving:
acceleration = (mass 1)(gravity) / (mass 1 + mass 2) = 19.6 / 7 = 2.8 m/s^2

Tension = 2.8 (5) = 14 N
or
Tension = -1 (2.8(2)-19.6) = 14 N
 

ChipB

PHF Helper
Jun 2010
2,367
292
Morristown, NJ USA
For part (ii) your answer would be correct if there is no friction, but there is.

I suggest you start by drawing a free body diagram for both bodies. For the 2Kg body you have forces of weight acting down and tension in the rope acting upward. For the 5 Kg mass you have tension pulling to the right and friction pulling to the left. Because both bodies are tied by a rope the magnitude of the tension acting on both bodies is equal, and the acceleration of both bodies is equal. You can set up an equation that equates the acceleration of both bodies - this will allow you to solve for T. Once you have that, you can solve for the acceleration.
 
Oct 2015
20
0
For part (ii) your answer would be correct if there is no friction, but there is.

I suggest you start by drawing a free body diagram for both bodies. For the 2Kg body you have forces of weight acting down and tension in the rope acting upward. For the 5 Kg mass you have tension pulling to the right and friction pulling to the left. Because both bodies are tied by a rope the magnitude of the tension acting on both bodies is equal, and the acceleration of both bodies is equal. You can set up an equation that equates the acceleration of both bodies - this will allow you to solve for T. Once you have that, you can solve for the acceleration.
Ah ok. I see where I went wrong.

ii) T - 4.9 N = 5 kg (a)
-T + 0.9800 N = 2 kg (a)
--------------------------------
-3.92 N = 7 kg (a)
-0.56 m/s^2 = a

T = 5 kg (-0.56 m/s^2) + 4.9 N = 2.1 N

That should be right now.

i) As for the first one, is it possible someone can give me some hints? I truly and honestly don't know what I'm doing. This is as far as I've gotten:

total mass = 7 kg
vf = 0 m/s
vi = ?
a = -0.56 m/s^2
Kinetic friction = 4.9 N

I don't have a formula I can think of to 100% use :( I don't know distance or time, so the Kinematics formulas don't look like they'll help me. I am absolutely stuck.

All I could muster was this:

Kinetic Energy = Kinetic Friction

Thus v = sqrt( Kinetic Fricition / .5(mass))
v = sqrt( 4.9 N / .5(7))
v = 3.74 m/s


Thanks for the help so far!
 

ChipB

PHF Helper
Jun 2010
2,367
292
Morristown, NJ USA
ii) T - 4.9 N = 5 kg (a)
-T + 0.9800 N = 2 kg (a)
I agree with the first equation, but not the 2nd. Where does the 0.98N term come from? I think it should be::

-T + W = ma --> -T + g x 2Kg = 2 Kg x a

T = 5 kg (-0.56 m/s^2) + 4.9 N = 2.1 N

That should be right now.
- No, I get a different answer.

i) As for the first one, is it possible someone can give me some hints?
Once you have the correct value for T, this is quite easy - just put T into one of the two formulas you started with and solve for 'a'. That's it!
 
Oct 2015
20
0
I agree with the first equation, but not the 2nd. Where does the 0.98N term come from? I think it should be::

-T + W = ma --> -T + g x 2Kg = 2 Kg x a



- No, I get a different answer.



Once you have the correct value for T, this is quite easy - just put T into one of the two formulas you started with and solve for 'a'. That's it!
Oh my goodness, what an error. I had been looking at an online guide for a little more help and put that number in instead of 19.6 N. My bad.

ii) T - 4.9 N = 5 kg (a)
-T + 19.6N N = 2 kg (a)
--------------------------------
14.7 N = 7 kg (a)
2.1 m/s^2 = a

T = 5 kg (2.1 m/s^2) + 4.9 N = 15.4 N

So... tension is 15.4 N and acceleration is 2.1 m/s^2.

That should be right now... for real this time haha.

Wait.... so it wasn't asking for speed/velocity???!?! It was looking for acceleration?!?!?!

Gee, now I feel so dumb >.<

On a side note, though, thank you so much ChipB for helping me with all my homework problems! It really means a lot! Thanks a million!
 

ChipB

PHF Helper
Jun 2010
2,367
292
Morristown, NJ USA
You are welcome! I must admit that asking for "speed" of an object that is accelerating is not appropriate - it should either ask for speed at a certain moment in time, or ask for its acceleration. You could say that the correct answer to that question is "impossible to answer with data provided."