How much force is used to pull the block?

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SUMMARY

The discussion revolves around calculating the force required to move a 7 kg block on a flat surface, transitioning from an initial velocity of 2 m/s to a final velocity of 4 m/s over a distance of 5 m. The coefficient of friction is given as 1. The net force calculated was 8.4 N, leading to the conclusion that the pulling force must exceed the frictional force of 68.6 N, resulting in a total pulling force of 77 N to achieve the desired acceleration.

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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!
 
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Dillion said:
F=ma
In that equation, what exactly does F represent? Be precise.
 
haruspex said:
In that equation, what exactly does F represent? Be precise.

Net force..?
 
Dillion said:
Net force..?
Yes. What actual horizontal forces are present here?
 
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)
 
i am sorry for posting a complete solution
It's just that you know I couldn't help myself
 
UchihaClan13 said:
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.
 
Ummm okay
I will
 
i edited it
Is it okay?
 
  • #10
UchihaClan13 said:
i edited it
Is it okay?
Yes , that's fine, thanks.
 
  • #11
You're welcome
 
  • #12
haruspex said:
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
Dillion said:
Force of friction = coefficient of friction * normal force
Yes.
Dillion said:
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 realize you have forgotten something.
Dillion said:
And the pulling force
Yes.

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

No. You should show units. A force is not just a number. If you fill in the units you might realize 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
Dillion said:
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
haruspex said:
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
Dillion said:
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
haruspex said:
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
Dillion said:
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
haruspex said:
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
Dillion said:
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
Yes it's 77 N!
You're correct!
:)
UchihaClan13
 

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