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My calculus text (Swokowski, Olnikc, Pence, 6th edition) gives the formula for work as $W = Fd$ and then goes on to explain that if the force varies over the distance the formula becomes an integral.

As part of an example, it then shows that the work to lift a 500 lb beam 30 feet would be 500 * 30 = 15,000 ft-lb. But don't we have to exert an upward force greater than the weight of the beam somewhere in our model in order to get the beam to move up? Then the work would be greater than 15,000 ft-lbs according to the definition of work.

It seems that in setting up integrals or just using the formula W = Fd examples always use the weight of the increment or object to be lifted without taking into account the fact that more than that force must be exerted at some point in order for the thing to move upwards.
By the formula W = Fd, if we greatly accelerate an object upwards, the work done lifting the object will be greater than if a lesser acceleration is applied across the same distance, but examples in my book don't seem to take this into account. (I am assuming that F = ma, and of course the mass stays constant.)

Qmechanic
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CandidFlakes
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  • Just so you know what is coming, there definition given here is a 1D restriction of the more general form $\mathrm{d}W = \vec{F} \cdot \mathrm{d}\vec{s}$ where $\mathrm{d}\vec{s}$ is a path element, and the '$\cdot$' represents the dot-product (AKA inner-product or scalar-product) between two vectors. – dmckee --- ex-moderator kitten Jun 20 '17 at 16:58

1 Answers1

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You are correct. To simplify matters, this amount is often ignored. There are several reasons why such a simplification is valid here.

  • In the first place, we have no minimum speed for the lift. By reducing the velocity, we can make the acceleration (and work needed to do so) as arbitrarily close to zero as we desire.

  • Any extra work done to accelerate can be returned during a deceleration. All that is required is that your path for the beam has it start and stop with the same speed. If the speed is the same, then kinetic energy at that point must be the same. That means any work done on the object must have gone into some other form and we assume it to be gravitational potential energy here.

BowlOfRed
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  • @NavneetKumar I'm not sure what you're asking about that situation. Perhaps you should ask it with some more detail as a new question. – BowlOfRed Jun 20 '17 at 18:25
  • Suppose,we throw a ball of mass,m upward with an acceleration, A such that A>gravitational acceleration (g).clearly,the ball must stop after reaching it's maximum height, h.so,HOW SHOULD THE POTENTIAL ENERGY HERE BE CALCULATED?,Here the definition of the potential energy proves wrong. Gravitational potential energy is defined as the amount of work done in raising it from the ground to that height,which is the weight of the object.I can prove this definition this way,The work done on the ball=F*displacement=mAh which is greater that the weight of the ball. Am I wrong somewhere,then please tell? – CandidFlakes Jun 20 '17 at 18:32
  • @NavneetKumar, I meant that this is no longer a comment on the original question. Please ask it as a new, separate question. Not a comment on this one. – BowlOfRed Jun 20 '17 at 18:36