0

I'm confused by an example in Stewart's (8th edition, early transcendentals) Calculus textbook. Example 1, Section 6.4: "How much work is done in lifting a 1.2kg book off the floor to put it on a desk that is 0.7m high?"

I can understand the computation: (mass) times (acceleration due to gravity) times (distance). What I don't understand is why the acceleration of the object as it gets lifted does not factor into the equation. I thought maybe we should just assume it's constant velocity, but that really doesn't make sense if the object is starting on the floor (where presumably the velocity is zero).

Is there a more precise way to phrase this question, like, "How much work is done against the force of gravity in lifting a 1.2kg book off the floor to put it on a desk that is 0.7m high?"

Is my confusion understandable? I'm not intentionally being pedantic. (I have a lot of respect for Stewart's Calculus book.)

Overall I was trying to understand (vector) line integrals better, and kept backtracking until I came to this example. I'm posting this in the math forum instead of the physics forum because I'm used to thinking about math definitions. (I last took a physics class in 11th grade.)

Thank you!

CJD
  • 420

1 Answers1

2

This is more like a physics question. But the answer is "It does not matter". If the book starts from rest and it ends at rest it is the same work if you accelerate constantly and decelerate constantly as if you accelerate just a little, stay constant speed, then slow down, or any other combination. This is because the gravitation is a conservative force, and the work done depends only on the initial and final positions (assuming that the velocities are the same - zero in this case)

Andrei
  • 37,370