How much force is used to pull the block?

  • Thread starter Dillion
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  • #1
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Homework Statement



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?

Homework Equations



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

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!
 

Answers and Replies

  • #3
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In that equation, what exactly does F represent? Be precise.

Net force..?
 
  • #5
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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
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i am sorry for posting a complete solution
It's just that you know I couldn't help myself
 
  • #7
haruspex
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i am sorry for posting a complete solution
It's just that you know I couldn't help myself
Please delete it now or I will have to report it.
 
  • #12
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Yes. What actual horizontal forces are present here?

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

And the pulling force
 
  • #13
haruspex
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Force of friction = coefficient of friction * normal force
Yes.
Force of friction = 1 * 7
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.
And the pulling force
Yes.

So how do these three relate: the frictional force, the pulling force, the net force that you calculated?
 
  • #14
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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?

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
haruspex
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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?
Yes, but you still need to mention the units. How much more?
 
  • #16
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Yes, but you still need to mention the units. How much more?


68.6 Newtons. I don't know how to calculate how much more....
 
  • #17
haruspex
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68.6 Newtons. I don't know how to calculate how much more....
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
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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?

The Frictional Force is -68.6N and the pulling force is positive N. The net force is 0?
 
  • #19
haruspex
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The Frictional Force is -68.6N and the pulling force is positive N. The net force is 0?
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
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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.

The net force was equal to 8.4 N.

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

Is the pulling force 77N?!
 
  • #21
haruspex
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The net force was equal to 8.4 N.

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

Is the pulling force 77N?!
Yes.
 
  • #22
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Yes it's 77 N!!
You're correct!
:)



UchihaClan13
 

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