Work done in lifting an object against gravity

ziaharipur

Dear fellows

I have three questions related to the topic “Lifting an object against gravity”.

If we lift an object of mass 5kg we need to apply a little more force than its weight to lift it and suppose we apply 50 N force (gravity is exerting 49 N forces on it) for just a second that produces 0.2 m/s/s acceleration in the body and then we reduce the force to 49N after a second, now according to Newton’s first law the body will keep moving upward with constant velocity (0.2 m/s). Am I right???????????

When we apply 50N force for a second this force does 0.1J of work on the body. Now during the lift we can’t say that the net work done is 0. Am I right??????????

When we make the body to be in rest position, we need to reduce the upward force for just 1N and as a result the net force will be downward and that will be 1N. Now this net force will do 0.1J of work in opposite direction or -0.1J and this will cancel out the 0.1J of work that was in upward direction. This is why we say that when we lift a book from table to shelf the net work done is 0. Am I right???????

russ_watters

Yes. I get 5 J, but whatever -- correct, the work for the rest of the lift is not zero. Yes. No, none of that is correct. Work against gravity is against gravity: that's the 49 N acting through whatever distance.

haruspex

After 1 sec it has risen 0.1m and reached 0.2m/s; so you've given it KE of 0.1J, but you've also increased its PE by 4.9J. Work done = 50N x 0.1m = 5J.
Well, you don't need to reduce it by that; that's just one way.
You're still applying an upward force, and the mass is still moving upwards, so you're still doing work on it. But since it is slowing, some of its KE is now going into its PE. I.e. continuing increase in PE = continuing work done on it by you + energy lost from KE.

ziaharipur

I posted this Question on yahoo answers in this way and got a very good answer showing that my understanding is correct


A book of mass 5kg is lifted from the table to shelf.

the body was at rest (initial kinetic energy=0), we applied 50 Newton force on it for just a second to produce an acceleration in the body, gravity is exerting 49N force on the body so the net force will be 1N upward this force produced an acceleration of 0.2 m/s/s in the body and after one second this force was reduced to 49 N (equal to the amount of force gravity is exerting on the body) now the body will keep moving in the upward direction with constant velocity that is 0.2 m/s (Newton’s First law of motion). this means that the kinetic energy of the body will be 0.1J and as W= change in Kinetic energy so the net work will be 0.1J during the lift (before coming in rest position on the shelf). When we reach the shelf we decrease our applied force to 48 N for just a second, in this way the net force will be 1N downward now this downward force will decrease the kinetic energy of the body from 0.1J to 0J in one second and the book will come in rest position. During the lift net work was 0.1J and after the lift both kinetic energies (initial and final) are zero so the net work is 0.

This is my understanding about lifting the book from table to shelf Am I right? If not then where I am wrong?


ANSWER
Sort of, depending on what you mean by "the lift". During the initial, acceleration phase, the net work on the book was 0.1J. During the constant-speed phase, the net work on the book was 0. During the deceleration phase, the net work was −0.1J. Add them all up, and the net work for the whole trip was 0.1 + 0 + (-0.1) = 0.

But remember that "Net work" (as in the equation W=ΔKE) means work due to ALL forces, including the positive work done by your hand and the negative work done by gravity. (Gravity does negative work because it exerts a force in the direction opposite from the motion.) If you wanted to calculate JUST the work done by your hand, it would be different, and would depend on the height of the shelf.