Calculating work done on a truck by friction when a truck is pushing a car

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Homework Help Overview

The discussion revolves around calculating the work done on a truck that is pushing a car, considering the forces involved, specifically the applied force and the opposing frictional force.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the calculation of work done, questioning whether the frictional force should be included in the work done on the truck. Some participants attempt to clarify the forces acting on the truck versus the car.

Discussion Status

There is an ongoing exploration of the correct interpretation of forces and their contributions to the work done. Some participants have provided alternative calculations and reasoning, while others have raised questions about the clarity of the original problem statement.

Contextual Notes

Participants note the potential ambiguity in the problem statement and the need for additional information to clarify the question being asked.

kristy hardy
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Homework Statement



A truck pushes a car by exerting a horizontal force of 500 N on it. A frictional force of 300N opposes the cars motion as it moves 4m. Calculate the work done on the truck.


Homework Equations



work = force x distance


The Attempt at a Solution



work done = 300N x 4 m = 1200J work right?
 
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kristy hardy said:

Homework Statement



A truck pushes a car by exerting a horizontal force of 500 N on it. A frictional force of 300N opposes the cars motion as it moves 4m. Calculate the work done on the truck.


Homework Equations



work = force x distance

For constant forces, the work would be:

[tex] W=F d\cos\theta[/tex]

where [itex]\theta[/itex] is the angle between the force diretion and the displacement direction.


The Attempt at a Solution



work done = 300N x 4 m = 1200J work right?

No, I believe that would be the magnitude of the work done by friction. Remember that if we want to calculate the work done on the truck, you have to use the forces that act on the truck.
 
My contribution: since the work done by resulting force on the body is the path integral along the path and this is a scalar (inner product from two vectors), is straightforward that the work of resulting force shall be the algebraic sum of work from each force acting on the considered body; in this way:

W = 500N x4 m - 300N x 4m = 800J

Danpos.
 
kristy hardy said:

Homework Statement



A truck pushes a car by exerting a horizontal force of 500 N on it. A frictional force of 300N opposes the cars motion as it moves 4m. Calculate the work done on the truck.


Homework Equations



work = force x distance


The Attempt at a Solution



work done = 300N x 4 m = 1200J work right?

Not correct: the 300N force is NOT the force acting on the truck.
 
The title of your post asks a different question that your post itself: which one is the actual question? And, don't you have more info?
 

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