3
$\begingroup$

I tried searching the answer to this all over the internet, but still cannot grasp this concept.

So, when a book is being lifted, let's say $25 meters$ with a mass of $4kg$, with constant speed. I have seen other similar problems, and the answer was done by using $mgh = 25* 4* 10 = 1000J$.

But, gravity would be doing negative work on the book as well in the opposite direction, right? So, it would be $-1000J$.

So, the object should not gain any potential energy, right?

Other equation I have been using to understand this is: $W = \Delta KE + \Delta PE$. So, $\Delta KE$ is zero, since no change in speed. And $W_{\rm net} = 0$? So, $\Delta PE$ should be zero, but it isn't, since there is an increase in height.

I know my thought is misled somewhere and would greatly appreciate as to why wouldn't the net work on this system not be zero.

$\endgroup$
  • $\begingroup$ First,you have to know what conservative force is. Suppose, in a gravity-less universe, you move the book upwards by applying force upto certain distance. Since, you have worked on it, it would have gained Kinetic Energy of the same amount, that you have lost. But when the gravitional force comes, it hinders the book's lifting, and part of the book's KE is stored as PE in the system. What's PE? It's just another energy! $\endgroup$ – user36790 Dec 11 '14 at 4:09
  • $\begingroup$ >I tried searching the answer to this all over the internet $\endgroup$ – Wolphram jonny Dec 11 '14 at 4:32
1
$\begingroup$

There are several possibilities for your confusion - and you bring up several related concepts (Work, potential energy, kinetic energy). When this happens I think it's helpful to isolate one thing which you know must, absolutely be true. The definition of work (for a constant force) is

$$W=\vec{F}\cdot \Delta \vec{x}=F\Delta x \cos\theta.$$

First ask "what is the work due to gravity?". Then $F=mg$, $\Delta x =x_f-x_i=$ 25 m (call this $h$), but the angle between $\vec{F}$ and $\Delta x$ is 180$^\circ$, so $cos(180)=-1$, and $W_g=-mgh=-1000$ J (the work due to gravity, as you've correctly identified).

In my view, this makes the most intuitive sense of all the things you've said. Force of gravity is acting to pull the object downward, so it's doing negative work on the object.

So, what now? We have various options for understanding the rest of your question. Let's try conservation of energy:

$$\Delta P+\Delta K=W$$

Ok, but when you write this equation, you need to know what "the system" is - so you can complete the statement, "I am looking for the change in potential and kinetic energy of the system". In this case, "the system" is the block, and the work being done on the block is the work by your hand. That's what makes the block go up, that's what adds energy to the system.

So, the work done by your hand is opposite the work done by gravity, and $W_h=mgh$. Now, in order to understand $\Delta K$, we need to know what the initial and final velocities are (which we have not used yet, or even needed). You've said they are both zero - fine, then we understand very well that

$$\Delta P=W_h=mgh$$

Slight aside: How can you tell when the change in potential energy is zero? Potential energy is how much energy an object "could use to do something". When you raise a book a certain distance, the book wants to move downward. Therefore, it has gained potential energy "to do something". This makes sense to me, but the definition is work is a bit more unambiguous so that's why I gave that first.

$\endgroup$
  • $\begingroup$ So, I was thinking it is like kinetic energy. You have to add all of "works" together to find the net change in kinetic energy. But, is that not the case with potential energy? Because, when you add Wh + Wg, it equals to zero. $\endgroup$ – kingkong5678 Dec 11 '14 at 5:31
  • $\begingroup$ @kingkong5678: Have you seen my links? I was also at blues regarding the concept. Read those links:) best of luck for your exam! $\endgroup$ – user36790 Dec 11 '14 at 5:44
  • $\begingroup$ kingkong5678: For a conservative force, doing work is just like changing potential. Since gravity is a conservative force, either is correct. However, when you lift something, gravity either does negative work on the system, or there is a positive change in the potential energy of the system. $\endgroup$ – levitopher Dec 12 '14 at 2:22
0
$\begingroup$

Doing work on an object will increase its kinetic energy, but it could also increase its potential energy and therefore not increase its $KE$. To raise a $4kg$ object by $25 meters$ you have to counteract the force of gravity. If your doing this with a force equal to $F_g$ then when you've lifted the object there will be no increase in $KE$.

To raise a $4kg$ object by $25 meters$ you have to do work. This work is the force you've applied across the distance you've displaced the object. Since there is no kinetic energy increase, the force you apply is equal to $Fg$ and therefore $W_{done}=mgh$

The object definitely gains $PE$. Consider what happens when you let it go at that height. Then gravity accelerates the object and you'll get an increase in $KE$.

Where did that energy come from? From the potential energy you helped the object gain as you raised it!.

$\endgroup$
0
$\begingroup$

Let us assume, you lifted the book infinitesimally slow and the velocity is zero so is the acceleration making change in kinetic energy zero but the work is never zero. It is mgh(m-mass of book, g- acc. of gravity, h-height to which the book is lifted).

The work done by you carrying the book around, that will be zero since no change in height or speed of the book with respect to you. With respect to a person who is static, the work is m/2 times v raised to the power of 2

$\endgroup$
0
$\begingroup$

You have misconcieved the expression for work in your question as you've assumed $W$ as the total work on the object. It is actually $W_g=\Delta K + \Delta {P.E}$ ,where $W_g$ is the net gravitational work on the object. You see, $W=0$ while $W_g=-mg\Delta h$ in your case. And thus, $\Delta P.E=-W_g=mg\Delta h$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.