# Force-work problem

1. Apr 19, 2010

### cperfetto91

A person drags a 30 kg box 8 m across a horizontal floor at a constant speed of 2 m/s. The coefficient of kinetic friction is 0.2 and the person pulls on the box at an angle of 30 degrees above the horizontal.
a) Draw a figure.
b) What is the total amount of work done on the box?
c) With what force is the person pulling on the box?
d) How much work does the person do on the box?
e) How much work does the floor do on the box?

W= Kef- Kei
F=MA
Ke= .5mv^2

2. Apr 19, 2010

### kuruman

What do you think? Did you do part (a) and draw a figure? What does the figure look like? What about part (b)? Can you figure the total amount of work done on the box using the relevant equations that you have posted?

3. Apr 19, 2010

### cperfetto91

I keep getting stuck on part C. The best attempt I made was that the force= Fn/sin30
and that Fx= Friction/cos30... Is that correct?

4. Apr 19, 2010

### kuruman

It is not correct. You cannot attempt part (c) unless you are sure that parts (a) and (b) are correct. So back to part (a). If you drew a figure,
1. What is the sum of all the horizontal components of forces and what is this sum equal to?
2. What is the sum of all the vertical components of forces and what is this sum equal to?

5. Apr 19, 2010

### cperfetto91

sum of the vertical= Fsin30
sum of the horizontal= Fcos30 - friction

(b) i got 60 J using w=Delta Ke

6. Apr 19, 2010

### kuruman

What happenend to the normal force?
What value did you get for the force of friction and exactly how did you get that value?
You got 60 J for the total work? What does the work-energy theorem say and what is ΔKE for an object that is moving at constant speed?

7. Apr 19, 2010

### cperfetto91

Normal Force and mg cancel out..... Mg=Fn
Friction=uFn .2(9.8* 30)= 58.8
Delta Ke= Kef-Kei=Work .5(30)(2^2)-0

8. Apr 19, 2010

### kuruman

That's not all you have in the vertical direction. You forgot to include the vertical component of the pulling force. Put that in and see what you get for the normal force. Then redo the force of friction.

Why is the initial kinetic energy zero? What does "constant speed" mean to you?

9. Apr 19, 2010

### cperfetto91

i see where you are coming from with the initial KE, but I dont understand, how else would you solve for it. And isn't the vertical component of the pulling for Fsin30

10. Apr 19, 2010

### kuruman

Solve for what? According to the work-energy theorem, the total work is the change in kinetic energy. Can you get a number for the change in kinetic energy? If "yes", then set the total work equal to that number.
Yes it is. So how would you write the sum of all the vertical components of forces? What is that sum equal to?

11. Apr 19, 2010

### cperfetto91

There is no change in ke because v is constant?

Fn-mg+Fsin30= ?

12. Apr 19, 2010

### kuruman

Correct.
What should you put in place of the question mark? What is the sum of all the vertical forces according to Newton's Second Law?

13. Apr 19, 2010

### cperfetto91

0 because the object is not moving in the y direction

So w=fDcos30

14. Apr 19, 2010

### kuruman

Correct for the sum of all the vertical forces.
How do you figure? The total work is change in kinetic energy and you just told me that the change in kinetic energy is zero. So what is the total work?

15. Apr 19, 2010

### cperfetto91

ooo, I'm sorry I thought since v is const. Change in Ke doesn't apply anymore, so total work is 0

16. Apr 19, 2010

### kuruman

You got it. I hope you can finish the problem because I am signing off for the time being.