Net work problem (attempted solution)

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Homework Help Overview

The problem involves calculating the work done on a 15.8 kg steamer trunk being dragged over a rough surface by a constant force of 90.9 N at an angle of 28.6 degrees above the horizontal. The trunk is moved a distance of 22.8 m, with a coefficient of kinetic friction of 0.18, and the acceleration due to gravity is 9.8 m/s².

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the calculation of work done by various forces, including gravity, normal force, pulling force, and friction. Some question the assumption that the work done by gravity and the normal force is zero, while others explore the components of the pulling force and their effects on friction.

Discussion Status

There are multiple interpretations of how to calculate the work done, particularly regarding the contributions of gravitational force and the components of the pulling force. Some participants have provided guidance on considering the vertical component of the pulling force and its impact on friction, but no consensus has been reached.

Contextual Notes

Participants note that the trunk is moving horizontally, which influences the calculations of work done by gravity and the normal force. There is also mention of the need to consider the distance in the work calculations.

tyger_eyes
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Homework Statement



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?

Homework Equations



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

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
 
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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.
 
tyger_eyes said:

Homework Statement



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?

Homework Equations



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

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

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?
 
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:
tyger_eyes said:
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)(22.8kg)cos28.6
=1819.637276 J
The work of friction is:
Wfr=Ffrxcosθ
=(27.8712)(22.8)(cos180)
=-635.46336

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...