# How much force is used to pull the block?

1. Jan 30, 2016

### Dillion

1. The problem statement, all variables and given/known data

A 7 kg block on a flat surface moves from 2 m/s to 4 m/s over 5 m. The coefficient of friction between the block and the surface is 1. How much force was used to move the block?

2. Relevant equations

Honestly, my physics instructor gave us so many worksheets with so many worksheets, I have no clue which one to use.

3. The attempt at a solution

vf^2 = vi^2 + 2a(change in x)

4^2 = 2^2 + 2a(5)
a=1.2
F=ma

F = 7*1.2
F=8.4

I don't know how friction comes into play though....

Thank you!

2. Jan 30, 2016

### haruspex

In that equation, what exactly does F represent? Be precise.

3. Jan 30, 2016

### Dillion

Net force..?

4. Jan 31, 2016

### haruspex

Yes. What actual horizontal forces are present here?

5. Jan 31, 2016

### UchihaClan13

Dillion,think for a minute would you
The net acceleration is 1.2 m/s^2
This acceleration can be yielded by applying a force F to the block which acts in the forward direction,opposing the frictional force(acting in the opposite direction)

6. Jan 31, 2016

### UchihaClan13

i am sorry for posting a complete solution
It's just that you know I couldn't help myself

7. Jan 31, 2016

### haruspex

Please delete it now or I will have to report it.

8. Jan 31, 2016

### UchihaClan13

Ummm okay
I will

9. Jan 31, 2016

### UchihaClan13

i edited it
Is it okay?

10. Jan 31, 2016

### haruspex

Yes , that's fine, thanks.

11. Jan 31, 2016

### UchihaClan13

You're welcome

12. Jan 31, 2016

### Dillion

Force of friction = coefficient of friction * normal force
Force of friction = 1 * 7

And the pulling force

13. Jan 31, 2016

### haruspex

Yes.
No. You should show units. A force is not just a number. If you fill in the units you might realise you have forgotten something.
Yes.

So how do these three relate: the frictional force, the pulling force, the net force that you calculated?

14. Jan 31, 2016

### Dillion

Ohhh it should be 1 *68.6 (7 * 9.8 = normal force). So the force of friction is 68.6.

"So how do these three relate: the frictional force, the pulling force, the net force you calculated?"

The net force is equal to 0 when the block is not moving. So the pulling force has to be more than 68.6, correct?

15. Jan 31, 2016

### haruspex

Yes, but you still need to mention the units. How much more?

16. Jan 31, 2016

### Dillion

68.6 Newtons. I don't know how to calculate how much more....

17. Jan 31, 2016

### haruspex

You calculated the frictional force as 68.6N. Let the pulling force be FP. Do those act in the same direction or in opposite directions? What is the net of those two forces?

18. Jan 31, 2016

### Dillion

The Frictional Force is -68.6N and the pulling force is positive N. The net force is 0?

19. Jan 31, 2016

### haruspex

You calculated the net force in your first post in this thread. It wasn't zero. If the net force were zero there'd be no acceleration.

20. Jan 31, 2016

### Dillion

The net force was equal to 8.4 N.

Frictional force (-68.6N) plus pulling force = 8.4 N.

Is the pulling force 77N?!