Help Finding the Applied Force to move a box?

In summary, a 4-kg box is moved from rest to a velocity of 6 m/s in 3 seconds by a certain force. The coefficient of friction between the box and the floor is 0.05. Using the equations for frictional force, net force, and final velocity, the acceleration is calculated to be 2 m/s squared. The net force is found to be 8 N, and the frictional force is calculated to be 1.96 N. To find the applied force, the net force and frictional force must be added, resulting in a final answer of 10 N.
  • #1
jisrael93
7
0

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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  • #2
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
PhanthomJay said:
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
jisrael93 said:
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
PhanthomJay said:
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
jisrael93 said:
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!
 
  • #7
PhanthomJay said:
Yes!

Thus making it 8 N plus 1.96 N, or 9.96 N...thanks a ton!
 
  • #8
jisrael93 said:
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
PhanthomJay said:
OK, good, but you should round off your answer to 10 N.

Right, to account for significant figures..got it.
 

1. How do I calculate the applied force needed to move a box?

To calculate the applied force, you will need to know the mass of the box, the coefficient of friction between the box and the surface it is resting on, and the acceleration due to gravity. You can then use the formula F=ma, where F is the applied force, m is the mass, and a is the acceleration. You will also need to take into account the direction of the force and any other external forces acting on the box.

2. What is the coefficient of friction and how does it affect the applied force?

The coefficient of friction is a measure of the resistance between two surfaces that are in contact with each other. It is dependent on the types of materials and the smoothness of the surfaces. A higher coefficient of friction means that there is more resistance between the two surfaces, so a greater applied force will be needed to move the box.

3. Can I use the same applied force for different boxes?

No, the applied force needed to move a box will vary depending on the mass and coefficient of friction of the box. A heavier box or one with a higher coefficient of friction will require a greater applied force to move it.

4. What are some ways to reduce the applied force needed to move a box?

One way to reduce the applied force is by using lubricants on the surface the box is resting on to reduce friction. Another way is to use wheels or other mechanical aids to assist with moving the box. Additionally, reducing the weight of the box can also decrease the amount of force needed to move it.

5. How can I ensure the safety of myself and others when applying force to move a box?

It is important to use proper lifting techniques and to not exceed your own physical limitations when applying force to move a box. It is also important to ensure that the box is stable and not at risk of tipping or falling. If working with others, communicate clearly and work together to safely move the box. If needed, use protective gear such as gloves or back braces to prevent injury.

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