# Net work problem (attempted solution)

1. Feb 24, 2010

### tyger_eyes

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

You drag a(n) 15.8 kg steamer trunk over a rough surface by a constant force of 90.9 Nacting at an angle of 28.6? above the horizontal. You move the trunk over a distance of22.8 m in a straight line, and the coefficient ofkinetic friction is 0.18.
The acceleration of gravity is 9.8 m/s2 .
How much is the work done on the block by the net force?

2. Relevant equations

Wnet=Wg+Wn+Wpull+Wfr
Wfr=Ffrxcosθ
Ffr=μFn

3. The attempt at a solution

Wg=mgxcos90=0
Wn=Fnxcos90=0

Wpull=Fpullxcosθ
=(90.9)(22.8)cos28.6
=1819.637276

Wfr=Ffrxcosθ Ffr=μFn=μmg
=(27.8712)(22.8)(cos180) =(.18)(15.8)(9.8)
=-635.46336 =27.8712

Wnet=Wg+Wn+Wpull+Wfr
=0+0+1819.637276-635.46336
=1184.173916

I have tried over and over and cannot get the right answer. If someone shows me where i went wrong...lifesaver

2. Feb 24, 2010

### themoose

The work done by gravity is not 0, it is only 0 if the direction of motion is horizontal. You need to find the component of gravity that is pulling the trunk down the slope.

Fg = mg is the force due to gravity, but the portion pulling it down the slope is small in comparison to the remainder of gravity pulling the trunk against the slope.

3. Feb 24, 2010

### Staff: Mentor

I'm not tracking how you are trying to solve it. Work = Force * Distance. So the work done is the force moving the trunk horizontally (the direction it is moving), multiplied by the distance it moves. The fact that the force is a vector acting up and sideways at an angle is significant for two reasons. What does the vertical component of the force do? What does the horizontal component do?

4. Feb 24, 2010

### tyger_eyes

The trunk is on a horizontal surface, it is being pulled at an angle of 28.6 above the horizontal.

I am attempting to sum all the work and sum it together. The Wg and Wn are both zero since the trunk is only moving horizontally.
The work of the pull is:
Wpull=Fpullxcosθ
=(90.9N)(22.8m)cos28.6
=1819.637276 J
The work of friction is:
Wfr=Ffrxcosθ
=(27.8712N)(22.8m)(cos180)
=-635.46336

Last edited: Feb 24, 2010
5. Feb 24, 2010

### Staff: Mentor

You forgot distance in "Wpull=Fpullxcosθ"

Work is force * distance.

The force that is pulling horizontally is opposed by friction. I wouldn't call it work done be frition -- that's not generally how you think about it. Rather, you do work as you pull against the frictional force.

Also, remember that the upward component of your pulling force does something to change the frictional force...

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