# Force and acceleration

1. Oct 29, 2006

### anderrod

Ok...so I am physics phobic! I read these physics questions and honestly have no idea what they are even asking. I have no idea how to start or what to do. I understand that this forum is for assistance for homework, but what do you do when you haven't a clue on how to even begin?

So I am looking for help on how you start with a problem like this:

You have a 10.0kg rock with a rope tied around it. What acceleration would you give the rock if you pull upwards on the rope with a force of 100 N?

How can I get an answer when I don't even know how far the rock traveled? Wouldn't I need to know that in order to answer the question?

2. Oct 29, 2006

### physicsrules

start out by drawing a free body diagram for the rock.. mg will be downwards and the 100 N will upwards. So the difference ( 100N - mg) will be equal to ma. and solve for a

3. Oct 29, 2006

### Tier

$$\vec{F}_{net} = \Sigma \vec{F} = m \vec{a}$$

Newton's Second Law should give you everything you need.

$$\Sigma \vec{F}$$
Would be the force the rock was pulled up with minus the weight of the rock
$$\Sigma \vec{F} = 100 - (10*(9.8))$$
then on the other side
$$m \vec{a}$$
would be the mass of the rock (10 kg) and then you could solve for a (acceleration)

4. Oct 29, 2006

### Ja4Coltrane

By the way, you asked about knowing distance. Acceleration could not care less for the distance something goes, acceleration is just how quickly something is changing speed and this is caused by being pushed. If you push something for a longer time, it's not going to change acceleration.
Does that help at all?

5. Oct 29, 2006

### anderrod

Thanks for the help, but I still am not getting it. I thought the answer was 10 m/s but when I submitted that answer, it was marked wrong. When I use your equations, I get .2 m/s and that just seems all wrong.

What am I not getting here?

I appreciate the help.

Thanks

6. Oct 29, 2006

### Tier

When something is being pulled (assuming the rope is massless) the force that it is being pulled with is not the only force acting on it, teh mass and gravity does work on the rock as well, making it harder to pull up with a faster acceleration. The sum of forces would then be equal to the force you pulled up with minus the Weight (in Newtons) of the object, so 100 - (10*9.8), and that would be equal to teh mass of the rock times its acceleration. Then you divide by the mass to get the acceleration

7. Oct 29, 2006

### geoffjb

Check your units. Acceleration is not measured in m/s.

8. Oct 29, 2006

### geoffjb

Except that the question asks for the applied force, not the net force.

9. Oct 29, 2006

### Tier

it has to be the net force though, becasue if you only factored in the applied force, you would get 10, which he said was wrong when he submitted it. In order to get the acceleration, Newton's Second Law has to be used, which implies net force

10. Oct 29, 2006

### anderrod

Ok...I have worked through the problem and I still am getting that same lame answer of .2 m/s^2. I get 2kgm/s^2=10kg*a
solving for acceleration, I get 2 kgm/s^2/10kg
which equals .2 m/s^2

So what am I still doing wrong?

Thanks. I see that before I neglected to account for the conversion to get the units right, but I still am not getting the right answer, so ????

Thanks for the assistance, I actually feel like I might make it through this problem.

C.

11. Oct 29, 2006

### geoffjb

An answer of $10.0 m/s^{2}$ (watch your sig figs and units) should not be wrong.

12. Oct 29, 2006

### Tier

Yeah, I guess that would be right, didn't read the question close enough

13. Oct 29, 2006

### anderrod

Hmmm....well now I am confused. I just checked with another student and they entered 10.0 m/s^2 and they got it wrong as well. I was pretty confident in my answer until I got it wrong. So now I don't know what to do. Thanks for the help anyway.

C.

14. Oct 29, 2006

### geoffjb

Might acceleration be negative upwards?