# Simple question about work and friction

Hi,
The simple definition of work done by a constant force on moving a constant mass is

(size of) force on object multiplied by (size of) distance object moves in direction of force while force is applied

In the ideal case of a puck on frictionless ice, an applied constant force will cause the puck to accelerate as long as the force is applied. The work done by the force on the puck transfers energy to the puck in the form of kinetic energy.

Ok – I understand that.

Take two scenarios of 1) pushing the puck on a rough surface and 2) lifting an object against gravity

In both case two balanced forces are acting on the object and the object (ideally) does not accelerate

Situation 1) the two forces acting on the puck are the pushing force and the friction. So which force is doing work on the block? Why, when we calculate the work done on the block do we use one force (say the pushing force) and ignore the other (friction)? Could we not argue that as the resultant force is zero then no work is done on the block? That won't work (pun intended) because we can't explain where the heat been the rubbed surfaces comes from.

Situation 2) is pretty much the same. We lift the object vertically against gravitational attraction of the earth. If the block does not accelerate the weight and lifting force are balanced and opposite. If the object is placed on a shelf at a height above the floor then it has gravitational potential energy so work must have been done on the block. Again how do we calculate the work done? Do we just multiply the lifting force by distance moved and ignore the weight? Yes, probably, but why?

Thanks

Clive

Hi,
The simple definition of work done by a constant force on moving a constant mass is

(size of) force on object multiplied by (size of) distance object moves in direction of force while force is applied

In the ideal case of a puck on frictionless ice, an applied constant force will cause the puck to accelerate as long as the force is applied. The work done by the force on the puck transfers energy to the puck in the form of kinetic energy.

Ok – I understand that.

Take two scenarios of 1) pushing the puck on a rough surface and 2) lifting an object against gravity

In both case two balanced forces are acting on the object and the object (ideally) does not accelerate

Situation 1) the two forces acting on the puck are the pushing force and the friction. So which force is doing work on the block? Why, when we calculate the work done on the block do we use one force (say the pushing force) and ignore the other (friction)? Could we not argue that as the resultant force is zero then no work is done on the block? That won't work (pun intended) because we can't explain where the heat been the rubbed surfaces comes from.

Situation 2) is pretty much the same. We lift the object vertically against gravitational attraction of the earth. If the block does not accelerate the weight and lifting force are balanced and opposite. If the object is placed on a shelf at a height above the floor then it has gravitational potential energy so work must have been done on the block. Again how do we calculate the work done? Do we just multiply the lifting force by distance moved and ignore the weight? Yes, probably, but why?

Thanks

Clive

In the two cases you quote, only one of each the two forces is acting in the direction of motion of the object.

jack action
Gold Member
Your problem is that you do not get the concept of work. Like you said, to get work, i.e. energy spent, you need a force and movement. If there is no motion there is no work done like the example of the book sitting on a shelf.

But to get the book on the shelf from, say, the floor, you need to spend energy which will be equal to weight of the book (the lifting force) times the distance between the shelf and the floor.

Similarly for the friction case, if there is no motion, there is no friction, hence no energy spent, no work done. As soon as the object moves, friction starts and the work done is equal to the friction force times the distance traveled.

For your puck on frictionless ice, you will input a force (= ma) for a certain distance (x) at which point the puck will have acquire a certain speed (v² = 2ax). At this point, if you remove the force, the puck will go at the acquired speed forever. In this case you have motion but no force, so there is no work done.

if there is no motion, there is no friction

Static friction?

jack action