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

Question

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?.

Answer 1

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,

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

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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.

Answer 2

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

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

Answer 3

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

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