# Homework Help: Pushing a box across horizontal surface

1. Oct 15, 2009

### Chandasouk

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

In a physics lab experiment, a 6.25 kg box is pushed across a flat table by a horizontal force F

If the box is moving at a constant speed of 0.400 m/s and the coefficient of kinetic friction is 0.120, what is the magnitude of F?

What is the magnitude of F if the box is speeding up with a constant acceleration of 0.160 m/s^2?

How would your answer to part (A) change if the experiments were performed on the moon, where g_m = 1.62 m/s^2?

How would your answer to part (B) change if the experiments were performed on the moon, where g_m = 1.62 m/s^2?

For part A, I found the weight of the block which was -61.25N and since it is on a horizontal plane Fn=mg, so Fn = 61.25N

I used the mew k being 0.120 and did

(.120)(61.25N) = 7.35N

What am I supposed to do with the constant speed it gives me? I know if something moves with a constant speed/velocity, the net force is zero...

2. Oct 15, 2009

### sermatt

The net force is the sum of all the forces acting on an object. In the case of constant velocity, the acceleration is 0, which says that there are NO unbalanced forces acting on the object.

Therefore, the applied force that pushes the box must be canceled out by the force of friction of an equal magnitude in the opposite direction.

Since you've calculated the force of friction, and Fnet = 0, and Fnet = Fapp - Ff, calculate the applied force!

3. Oct 15, 2009

### Chandasouk

Applied force would then be 7.35N.

Okay, onwards to the second part. What is the magnitude of F if the box is speeding up with a constant acceleration of 0.160 m/s^2?

Would I do

F = ma

F=(6.25)( 0.160 m/s^2) = 1N ?

For some reason, I think that is incorrect.

4. Oct 15, 2009

### sermatt

If the box is accelerating, its net force would be greater than zero.

In this case, you're right, Fnet=(6.25)( 0.160 m/s^2) = 1N.

Therefore, the NET force is 1N. However, net force also means that there is an unbalanced force acting on it, so 1N = Fapp - Friction (because since the object is accelerating, the applied force must be greater).

Since you know the value of friction, sub it into the equation 1N = Fapp - Ff.

EDIT: To find the value of the applied force.

5. Oct 15, 2009

### Chandasouk

Okay, with that clarified. The next two seem easy?

How would your answer to part (A) change if the experiments were performed on the moon, where g_m = 1.62 m/s^2?

The weight of the object would be (6.25)(1.62)=10.125N

Frictional Force would be (.120)(10.125)=1.215N

Thus,

Fnet = Fapp - Ffr

0=Fapp-1.215N

Fapplied = 1.215N

and that is our answer for part C

Part D

F=ma

F=(6.25)( 0.160 m/s^2) = 1N

1N = Fapplied - Ffr

1N=Fapplied - 1.215N

Fapplied = 2.215N

6. Oct 15, 2009

### sermatt

Precisely.

All you have to do now is understand this concept 100% so you can find your way out of more sophisticated questions.

7. Oct 16, 2009

### Chandasouk

I guess the main thing I have to remember is F =ma is for Fnet which is the sum of all forces. In this particular problem, the verticxal axes cancel each other out so the only forces acting are the horizontals which consist of Fapplied and F-friction. Therefore i can set up like so

ma=Fapplied-Ffriction