Net work problem (attempted solution)

In summary, the work done on the block by the net force is 1184.173916 J. This is calculated by summing the work done by the pulling force, which is 1819.637276 J, and the work done against friction, which is -635.46336 J. The work done by gravity and the normal force are both negligible in this scenario since the trunk is moving horizontally. It is important to consider the vertical and horizontal components of the pulling force and how they affect the frictional force in order to accurately calculate the net work done.
  • #1
tyger_eyes
2
0

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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  • #2
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
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?
 
  • #4
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:
  • #5
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...
 

1. What is a net work problem?

A net work problem is a type of mathematical problem that involves finding the work done by an object or system when it is subjected to various forces. It is commonly used in physics and engineering to calculate the amount of energy required to move an object or perform a task.

2. How do you solve a net work problem?

To solve a net work problem, you will need to use the formula W = Fd, where W represents work, F represents force, and d represents displacement. You will also need to know the direction of the force and displacement, as well as any other relevant information such as the mass of the object.

3. What are some common mistakes when solving a net work problem?

One common mistake when solving a net work problem is forgetting to convert units. Make sure that all units are consistent and in the correct form before plugging them into the formula. Another mistake is not considering the direction of the force and displacement, which can result in an incorrect answer.

4. Can you use the net work formula for all types of forces?

No, the net work formula can only be used for constant forces. If the force is not constant, the work done will need to be calculated using calculus. Additionally, the net work formula does not take into account other factors such as friction or air resistance, which may affect the overall work done.

5. How is net work related to other concepts in physics?

Net work is closely related to other concepts in physics, such as energy and power. Work is a form of energy, and the net work done on an object is equal to the change in its kinetic energy. Power is also related to work, as it measures the rate at which work is done. Net work is an important concept in understanding the movement and behavior of objects in the physical world.

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