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

[kristy hardy]

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?

 

[alphysicist]

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:

<br /> W=F d\cos\theta<br />

where \theta 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.

 

[danpos]

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.

 

[borgwal]

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.

 

[borgwal]

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?