# How can I find the mechanical energy at a given point when there is dissipation of energy during fall?

The problem is as follows:

A sphere of $$2\,kg$$ is released (assume $$g=10\frac{m}{s^2}$$) from a height of $$2\,m$$ with respect of a horizontal surface. If due air resistance $$10\,J$$ are dissipated by each meter that the sphere goes down. Find the mechanical energy with respect to the ground (in Joules) on the instant when the kinetic energy is equal to $$0.5$$ times its potential energy.

The alternatives given are as follows:

$$\begin{array}{ll} 1.&350\,J\\ 2.&300\,J\\ 3.&250\,J\\ 4.&200\,J\\ \end{array}$$

I'm confused exactly how to tackle this problem.

Essentially what I believe is that:

$$E_{k}=\frac{1}{2}E_{u}$$

But the thing here that comes into play is that there's energy dissipated when the sphere is falling.

Then the mechanical energy would be as follows:

$$E_{u}=E_{k}+E_{u'}$$

At that given height there would be kinetic energy and potential energy.

But it mentions that the kinetic energy will be at that given height $$\frac{1}{2}E_{u}$$

$$E_{u}-150=\frac{1}{2}E_{u}+E_{u'}$$

$$mg(2)-150=\frac{1}{2}(mg(2-x))+mg(2-x)$$

By inserting the given values for the mass and gravity this is reduced to:

$$x=\frac{17}{3}$$

But that's where I became stuck. Can someone help me with this problem?.

• What is $E_{u'}$? Where did the number $150$ come from? Jan 28 '20 at 14:07
• @DavidK Sorry for that. It was part of my attempt into solving the problem. In the rush I thought that the energy dissipated was $150\,J$ but it was $10\,J$. But overall it isn't right. Jan 29 '20 at 13:51
• but .. the starting potential energy $mgh= 2\cdot 10 \cdot 2 = 40 J$, isn't it ? Jan 30 '20 at 18:19
• all options are incorrect object would never have more energy than it had at the start (40 J in the form of gravitational potential. energy)
– user454960
Jan 30 '20 at 20:20

Here we are dealing with kinetic energy $$E_k$$, potential energy $$E_u$$, and energy transfered to the air by the effect of "air resistance"--let's call this $$E_f$$. We might be tempted to express each of these as a function of time, but since the sphere passes through each height at a unique time we can alternatively express each kind of energy as a function of height $$x$$, where $$x=2$$ at the point of release.

Energy is preserved so, the sum is constant:

$$E_k(x_1) + E_u(x_1) + E_f(x_1) = E_k(x_2) + E_u(x_2) + E_f(x_2)$$

where $$x_1$$ and $$x_2$$ are any two heights. Try setting $$x_1 = 2$$ and let $$x_2=h$$ be the height at the instant when the kinetic energy is equal to $$0.5$$ times its potential energy, as asked.

$$E_k(2) + E_u(2) + E_f(2) = E_k(h) + E_u(h) + E_f(h). \tag1$$

Make formulas for as many of these as you can, either numbers or functions of $$h.$$

You should be able to get $$E_k(2)$$, $$E_u(2)$$, $$E_f(2)$$, $$E_u(h)$$, and $$E_f(h)$$ directly from the problem statement, and you can express $$E_k(h)$$ in terms of $$E_u(h)$$ based on the "$$0.5$$ times" condition, so now you have $$E_k(h)$$ as a function of $$h.$$

Plug everything you have into Equation $$(1)$$ and you'll have an equation with just one unknown, $$h.$$ Solve it.

Now you know $$h$$ so you can compute the answer,

$$E_k(h) + E_u(h).$$

An alternative approach: In the absence of air resistance, kinetic energy would increase by $$20$$ J for each meter the sphere descends. But we lose $$10$$ J to air resistance for each meter. Since $$20 - 10 = 10,$$ the kinetic energy increases at a rate of only $$10$$ J per meter.

On the other hand, potential energy, which starts at $$40$$ J, still decreases at the rate of $$20$$ J per meter. That is, the loss of potential energy is twice the gain of kinetic energy (since $$20/10 = 2$$):

$$E_u = 40 - 2 E_k.$$

Knowing that $$E_k = \frac12 E_u$$ at the instant the problem wants you to find the energy, find $$E_u$$ at that instant, find $$E_k$$, and add them.

• Why should I account for $E_f(2)$ as if there's air resistance at very instant when the ball is released if there isn't any motion?. If so $E_f(2)=10$? Jan 29 '20 at 13:56
• $E_f(2)$ is not a quantity of force, it's a quantity of energy. Specifically, it's the amount of mechanical energy that has been lost by the sphere since it was released. At the instant that the sphere is released, at $x=2$, how much mechanical energy has the sphere lost since the time it was released? That should tell you what value to assign to $E_f(2).$ Jan 29 '20 at 14:02
• By looking on the answers sheet it says that the answer to this problem is $300\,J$ this doesn't make any sense to me. But I can understand that right at the beginning there is no motion, no friction acting on the ball, just potential energy. Therefore $E_{k}=E_f(2)=0$ are we right on this?. Using the whole condition thing I'm arriving at $h=1\,m$. The thing is the interpretation of mechanical energy. Jan 30 '20 at 15:34
• In this case I believe they're asking for the contribution of the kinetic energy and the potential energy alone (without the frictional part) which it would be $30\,J$. Because it it would be only potential energy that would make $20\,J$ and if they would be asking for all energy it would be $40\,J$ as it is preserved all the way down and there wouldn't be any bother to do any rest of calculations other than finding the potential energy at the top. Am I right with this logic?. Can you explain if I am good with this to me please?. Jan 30 '20 at 15:38
• Based on the numbers you have, I also get 30 J at a height of 1 m as the solution. Your book seems to have an issue with the placement of the decimal point, the answers are all off by a factor of 10. If we started at 20 meters height then the answer would be 300, so maybe they dropped a digit from the height. Jan 30 '20 at 17:46

Attempt:

Let $$y=0$$ at the ground, $$y=h=2$$ at point of release.

After $$(h-y)$$ meters fall $$E_{diss} =-10(h-y)$$ energy lost.

$$E_{pot}(h)=mgh$$;

$$E_{pot}(h)= mgh=$$

$$(1/2)E_{pot}(y) +E_{pot}(y) +10(h-y);$$

$$E_{pot}(h)=2(10)2=$$

$$(3/2)2(10)y+10(2-y);$$

$$40=30y+20-10y$$;

$$20y=20$$; $$y=1$$.

$$E_{pot}(y=1)=2(10)(1)=20$$.

Note:

0) Without dissipation

Energy conservation:

At any point $$y$$ above ground we have

$$E_{kin}(y)+E_{pot}(y)=E_{pot}(h)=mgh;$$

1) With dissipation:

Energy lost falling a distance $$(h-y)$$:

$$E_{diss}(y)=-10(h-y)$$;

The remaining energy is split :

$$E_{kin}(y)+E_{pot}(y)=$$

$$E_{pot}(h)-10(h-y)$$;

Additional info: For some $$y$$ we have

$$E_{kin}(y)=(1/2)E_{pot}(y)$$.

• Apparently there isn't a match within the alternatives given. Can you explain the reasoning behind the equation which you have used?. I know it is the conservation of mechanical energy but if energy is dissipated why is it $+10(h-y)$? and not $-10(h-y)$?. Jan 29 '20 at 13:54
• I try.Start with the given energy. E=mgh, at h above the ground. At some height y (>0) above the ground you have E_pot =mgy ; E_kin=(1/2)mgy (given) now add energy that has been lost : 10(h-y)(here positive), to mgy +(1/2)mgy +10(h-y)= mgh to recover, equate to the original energy. Yes, dissipated energy is lost. Hence E_kin is smaller, adding the "lost part" gives you the energy balance at t=0.Your thoughts? Jan 29 '20 at 14:13
• I'm confused exactly on why the answer should be $20\,J$ and not $40\,J$ haven't we established that the mechanical energy (which is what they're asking at that point be the same at what it was just at the beginning when the ball was released?. I mean accounting for the dissipation due air resistance. In this case you're only calculating the potential energy for that given height. Why? shouldn't account also for the kinetic energy?. Jan 30 '20 at 15:29
• Can you please explain this part?. Or is it that what they are asking is the energy without the disspation which it would be $30\,J$?. The answers sheet says that the answer is $300\,J$ but this doesn't make any sense, I suspect that there's an added zero by error. But if judging from this perspective I believe that the answer would be $30\,J$. Jan 30 '20 at 15:32
• Chris.Start with potential energy mgh (h=2).If there is no dissipation E(kin)(y), kinetic energy at y above ground, + E(pot)(y). above ground =mgh .Correct? The sum of them equals the original E(pot)(h). Now energy is lost -10(h-y), y location above ground (h=2, and 0\le y\le 2. At any point y above ground we have left E(pot)(h)-10m(2-y) to be "shared between E(kin)(y) and E(pot)(y). I edit , give me a few moments. Jan 30 '20 at 17:38

we know ,by work-energy theoram

$$(U_{f}-U_{i})+(K_{f}-K_{i})=\Delta U +\Delta K=W_{nc}=\Delta E$$

where $$E$$ is total mechanical(kinetic $$K$$ + potential $$U$$) energy at any instant & $$Wnc$$ is work done by non conservative forces,air drag ,at same instant. chose ground as reference for potential energy (that is $$U_{f}=0$$) where subscript $$f$$ means final position

apply WETheo. between top point and ground to get final kinetic energy of sphere when it hits ground

$$(0-20\times2)+(K_{f}-0)=-10\times2\implies K_{f}=20 J$$

again apply WE theo. bewteen any intermediate point and ground (consider ground as final position and intermediate point as initial)

$$[0-20(2-y)]+\left[20-0.5 (20(2-y))\right]=-10(2-y)\implies y=1$$

so, total mechanical energy at this instant

$$20(2-1) + 0.5(20\times (2-1))=30J$$