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

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

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.

2. Relevant equations

work = force x distance

3. The attempt at a solution

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

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 Quote by kristy hardy 1. The problem statement, all variables and given/known data 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. 2. Relevant equations work = force x distance
For constant forces, the work would be:

$$W=F d\cos\theta$$

where $\theta$ is the angle between the force diretion and the displacement direction.

 3. 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.

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

 Quote by kristy hardy 1. The problem statement, all variables and given/known data 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. 2. Relevant equations work = force x distance 3. 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.