Help Finding the Applied Force to move a box?

  • Thread starter jisrael93
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  • #1
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Homework Statement


A certain force moves a 4-kg box from rest to a velocity of 6 m/s in 3 seconds. If the coefficient of friction between the box and the floor is 0.05, what is the magnitude of the force?
Given/Known:
mass = 4 kg
μ (I'm assuming it means kinetic friction?) = 0.05
time = 3 seconds
initial velocity = 0 m/s
final velocity = 6 m/s
gravity = 9.8 m/s squared


Homework Equations


Frictional Force = μ(mg)
Net Force = (m)(a)
final velocity = (initial velocity)(acceleration x time)


The Attempt at a Solution


From these equations, I calculated the acceleration to be 2 m/s squared. This is where I get confused - if you find the net force with the equation mass x acceleration, then you get 8 N for a net force. I calculated the frictional force with the equation above, so it was like this: (0.05)(4)(9.8) and I got an answer of 1.96 N. The natural force on the box is found by multiplying mass times the force of gravity, which ends up being 39.2 N. Now, my question: do I subtract the frictional force of 1.96 N (if that is right) from the natural force of 39.2 N to get the answer I'm looking for? Does the net force even get used in this problem? We only did a problem to find the acceleration with a given force, but not the other way around. Thanks to anyone that can help me!
 
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Answers and Replies

  • #2
PhanthomJay
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You have to look in the horizontal and vertical directions separately. You have correctly solved for the acceleration and net force in the horizontal direction. So if the net force is 8 and the friction force is about 2, then the applied force must be what? Note that the net force is always in the direction of the acceleration , and that the friction force opposes the direction of relative motion. Note also that in the vertical direction, there is no acceleration.
 
  • #3
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You have to look in the horizontal and vertical directions separately. You have correctly solved for the acceleration and net force in the horizontal direction. So if the net force is 8 and the friction force is about 2, then the applied force must be what? Note that the net force is always in the direction of the acceleration , and that the friction force opposes the direction of relative motion. Note also that in the vertical direction, there is no acceleration.
So, would I be correct in subtracting the friction force of 1.96 N from the net force of 8 N, since they oppose each other?
 
  • #4
PhanthomJay
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So, would I be correct in subtracting the friction force of 1.96 N from the net force of 8 N, since they oppose each other?
No, the net force is the algebraic sum of the forces, that is, Fnet = Fapplied plus Ffriction, watch plus and minus signs .
 
  • #5
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No, the net force is the algebraic sum of the forces, that is, Fnet = Fapplied plus Ffriction, watch plus and minus signs .
Ah, I think I understand. If Fnet = Fapplied plus Ffriction, and Ffriction opposed the Fnet making it negative, then that would make the equation Fnet = Fapplied minus Ffriction, so the Fapplied would be Fnet plus Ffriction...is that correct?
 
  • #6
PhanthomJay
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Ah, I think I understand. If Fnet = Fapplied plus Ffriction, and Ffriction opposed the Fnet making it negative, then that would make the equation Fnet = Fapplied minus Ffriction, so the Fapplied would be Fnet plus Ffriction...is that correct?
Yes!
 
  • #8
PhanthomJay
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Thus making it 8 N plus 1.96 N, or 9.96 N...thanks a ton!
OK, good, but you should round off your answer to 10 N.
 
  • #9
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OK, good, but you should round off your answer to 10 N.
Right, to account for significant figures..got it.
 

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