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A block of mass $2kg$ is kept at origin at $t=0$ and is having velocity $4{\sqrt 5}$ m/s in positive x-direction. The only force on it is a conservative and its potential energy is defined as $U=-x^3+6x^2+15$ (SI units). Its velocity when the force acting on it is minimum (after the time $t=0$) is

And the Solutions is:

At $x=0$

$K= \frac{1}{2} (2)(80)$ and $U= 15J$

Total energy is, $E=K+U=95J$

Force,$ F= \frac{-dU}{dX}$

$F= 3x^2-12x$

for F to be minimum, $ F= \frac{-dF}{dX}=0$

$x=2m$

At $x=2m$

$E=K+U$

$95= \frac{1}{2}(2)(v^2) + (-8+24+15)$ $v=8 m/s$

Now my doubt is why $ F= \frac{-dF}{dX}$ needs to be zero for minimum force?

I know I have some doubt in my basic concept so it would be very helpful if you'll be elaborate.

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  • $\begingroup$ I think you are mistaking a mathematical minimum with a physical one. For kinematic and dynamic quantities, the negative sign indicates direction. So although the mathematically, the minimum should be at the vertex of the parabola force by the graph of Force, physically, the minimum possible force is 0. Here, we are specifically concerned with the magnitude Of the force $\endgroup$ – OVERWOOTCH Jul 30 '20 at 17:03
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"Now my doubt is why F=−dF/dX needs to be zero for minimum force?"

Note, F is not equal to -dF/dx, you probably got confused with U = -dF/dx

here, we say we have minimum value of force when dF/dx =0

dy/dx is used to represent slope at a specific point, it could be any point over where a function is continuous.

now if we say something is minimum, all the values around it should be greater than the minimum value. in this case, force is taken in y axis and distance 'x' is taken in the x axis

so, the slope for all the points before the point of minimum force, will be negative as the values of force decrease till they reach the minimum value. and when it crosses the minimum point, values of force start increasing i.e the slope becomes +ve.

but, if we look at the graph, it doesn't have a point where there is sudden transition from -ve slope to +ve slope, hence there has to be a point where slope is 0, which will be the minimum value as there the value neither increases or decreases, that will be the minimum value as all the values arount it are more than that.

hence we say, in this case, force is minimum when df/dx is 0

this is a very simplified explanation, it can be understood better after calculus is done in mathematics. still, if you are interested, there is a very intuitive explanation by a channel 3 Blue 1 Brown on youtube, give it a shot if you are intersted.

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Since F is a function of x, and it's an upward parabola the minimum value is obtained at the vertex of the parabola where the slope of the curve is 0. You will study this in calculus later on.

enter image description here

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for F to be minimum, $F=−\frac{dF}{dX}=0$

Watch out, this is not correct. I think you mixed two different formulas, the relatin between force and potential $F=-\frac{dU}{dx}$ and the condition to be a minimum $\frac{dF}{dx}=0$. If your question was why this is the condition for being a minimum, there are several ways to see this. For me, the most intuitive one is thinking about the meaning of the derivative, starting from the very basic incremental ratio definition (Note: my mother language is Spanish, "incremental ratio" is the direct translation from there, but I'm not sure it's a term used in English as well):$$F' = \lim_{\Delta x \to 0} \frac{F(x+\Delta x)-F(x)}{\Delta x}$$ A positive value for $F'$ means that if $\Delta x>0$ (you move to the right), $F$ grows on that direction, while if its negative (you move leftwards) it decreases. A (local) minimum of $F$ is defined as a point which is smaller than all the other points nearby, so if you approach this minimum point, $F$ decreases, and if you go away from it, $F$ increases. If, like in your problem, the minimum is located in $x=2m$, if you start from $x=1m$ and approach it, $F$ decreases (so $F'$ is negative), and once you reach it, if you go further, to $x=3m$ let's say, F increases again ($F'$ is positive). So, you started from $x=1m$, where $F'$ was negative, and ended in $x=3m$, where $F'$ is positive, so using Bolzano–Weierstrass theorem it must become zero at some point. You can approach the minimum as long as you want, and this relation is still valid, so the minimum is the point where the derivative becomes zero.

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First you need to understand how the potential energy is changing with $x$ you will see that given parabola will see a increase in a potential energy after crossing that point from both the side ,there will be decrease in the potential energy of the particle till it reaches that point (you need to find point by yourself) it means the particle is moving towards the field so there will.be decrease in the potential energy so you have asked to find

the velocity at the point where force is minimum. So velocity will be minimum at this point X which we have find above. Note;the graph is similar to shm graph

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It depends on how you want to define minimum F. You might say it occurs when F = 0 (at x = 4) or at the lowest point on a plot of F vs x which occurs when the slope of the curve is zero (F is -12 at x = 2).

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This is because as you can see in the solution, $$F= 3x^2-12x$$, i.e $F$ is dependent on x. Now $\frac{dF}{dx}$ is basically the slope of the $F$ vs x graph. Now, for any continuous graph as in this case, at the maximum or the minimum of the graph, the slope will be 0. Here, since the minimum force is required, the slope and hence $$\frac{dF}{dx}=0$$ Moreover, it's $\frac{dF}{dx}$ and not $-\frac{dF}{dx}$. I think you have mixed it up with the formula $F=-\frac{dU}{dx}$. I hope it helps you.

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