TagEqual force for different times: Pushing a car slowly vs quickly

[thetexan]

[Mentor's note; thread title edited to be more descriptive]
If I push a 2000 lb car 100 feet slowly across the parking lot say taking 10 minutes to do it, and later do it again this time taking 2 minutes to do it, will I have used approximately the same amount of total force to accomplish both tasks?

In other words, does less force applied over a longer period equal the force applied over a shorter period if the same task is performed...that is move 2000 lbs over 100 feet?

tex

 

[Nugatory]

[Mentor's note; thread title edited to be more descriptive]
If I push a 2000 lb car 100 feet slowly across the parking lot say taking 10 minutes to do it, and later do it again this time taking 2 minutes to do it, will I have used approximately the same amount of total force to accomplish both tasks?

In other words, does less force applied over a longer period equal the force applied over a shorter period if the same task is performed...that is move 2000 lbs over 100 feet?

A smaller force over a longer distance can do the same amount of work as a larger force over a shorter distance; this follows from the basic $W=Fd$ relationship between force, distance, and work done.

However, there's more to your car problem than this. First you have to apply force to the car to accelerate it to some speed; but once it's started moving it will coast at that speed, and the only force you'll be applying is to overcome friction and rolling resistance in the tires. Finally, at the other end of the parking lot the car is still moving, and of course it's going faster if it crossed the parking lot in two minutes than in ten minutes. That greater speed means that you either applied a greater force to get it moving, or you applied the same force over a longer distance, or both. So you'll have to look at the actual forces and distances through both the initial acceleration and coasting against friction stages to calculate the work done. You will almost certainly find that the work done is very different in the two cases. The difference will be found in energy lost to friction and in the kinetic energy of the car when it reaches the end of the parking lot and is still moving.

As an aside: If there were no friction, and if you stopped the car at the end of the parking lot using an ideal regenerative braking system that recovered and stored (typically, by charging a battery) the kinetic energy of the car, you would find that there was no net work done. The energy that you initially expended to start the car moving would all end up stored in the regenerative braking system for future use.

 

[jbriggs444]

In other words, does less force applied over a longer period equal the force applied over a shorter period if the same task is performed...that is move 2000 lbs over 100 feet?

To a good approximation it takes the same continuously applied force regardless of whether you take two minutes or ten. You will be supplying enough force to overcome friction either way.

"Total force" is not a term that would be used to describe this.

One would use "impulse" to describe the product (or integral) of force applied times duration. This gives the "momentum" imparted to the car. Of course, friction will be acting in the opposite direction for the same interval. So the car gains little or no momentum as a result of the applied force.

One would use "work" to describe the product (or integral) of force applied times distance moved in the direction of the force. This gives the "energy" imparted to the car by the force. Of course, friction will again be acting in the opposite direction over the same path. So the car gains little or no energy as a result of the applied force".

One would use "power" to describe the rate at which "work" is done on the car. The faster you move the car with a given force, the more higher the rate of energy expenditure -- the more "power" is being applied.

None of this reflects directly on how much effort you feel you have put in, how tired you feel or how sweaty you become. Those are matters involving the internal efficiency of your human muscles.