0
$\begingroup$

So, we have a body $P_1$ of mass $m$ that travels from $A$ to $B$ along a straight line, on wich is exerted a constant force $F$ dependent only on the position of $P_1$. We know that the work done is $W_{P_1} = \frac{1}{2}m(v_{1A}-v_{1B})^2$, with the expression within the parenthesis being the difference in velocity from $A$ to $B$. Now, suppose that a second body $P_2$ with the same mass $m$ travels from $A$ to $B$. We know that the work depends only on the path and the force, so the work $W_B$ on $B$ satisfies $W_B = W_A$, hence $\frac{1}{2}m(v_{1A}-v_{1B})^2 = \frac{1}{2}m(v_{2A}-v_{2B})^2$ and $(v_{1A}-v_{1B})^2=(v_{2A}-v_{2B})^2$. Assuming a force opposite to the direction of motion, we conclude:$v_{1A}-v_{1B}=v_{2A}-v_{2B}$, in other words the change in velocity over the path $AB$ of both the bodies does not depend on their initial velocity. But I know this is false: is there a fallacy in my logic? In my math? Where am I doing a mistake?

$\endgroup$
2
  • 2
    $\begingroup$ Your equations are wrong. The work is equal to the change in kinetic energy. What you wrote is not the difference in kinetic energy. Fix this and then reformulate the question. $\endgroup$
    – nasu
    May 10 at 1:37
  • 4
    $\begingroup$ $(x-y)^2 \neq x^2 - y^2$ $\endgroup$ May 10 at 1:41

2 Answers 2

0
$\begingroup$

We know from Newton’s second law that $\vec{F}=\frac{d\vec{p}}{dt}$. If the mass of the particle/object/body is constant, we can write $\vec{F}=m\frac{d\vec{v}}{dt}$. Rearranging the equation gives $d\vec{v}=\frac{\vec{F}dt}{m}$. This means that the difference in velocity only depends on the force applied on the body, the time it’s applied for, and the body’s mass. According to the work-energy theorem, the work done on or by a system is equal to the change in it’s kinetic energy. If you apply a large(r) force, over some distance d, the change in velocity $d\vec{v}$ will be large(r), and hence, the difference of the squares of the final and initial velocities will be large(r). Similarly, applying a small force over the same distance d, the change in velocity will be small(er), and the difference between the squares of the final and initial velocities will be small(er). Also, note that you have made a mathematical error in your expression for the work done. Since the work done is the difference in kinetic energy, $W=\frac{m}{2}(v_f^2-v_i^2)$ and not $\frac{m}{2}(v_f-v_i)^2$ as you’ve written.

$\endgroup$
0
$\begingroup$

$$F= m_{0}a$$

In classical mechanics, if there is the same force, then at any speed the acceleration is going to be the same.

However, there is only the same change in momentum if $\int \vec{F} \cdot dt$ is the same.

For a body in the presence of a force field , moving through a distance x, regardless of speed, the body gains some kinetic energy KE.

$KE \propto v^2$

enter image description here

I have taken the libery to draw lines indicating the kinetic energy of a body of 1, 2 and the corresponding velocities.

Let's say a body is at at rest, and gains "1" of kinetic energy, because KE is proportional to v^2, the total change in velocity is 1m/s

Now, let's say the body is ALREADY moving at 1m/s, once it travels through a distance X under a force field it gains "1" KE such that it is now has 2 Ke,

The change of kinetic energy is the same for both, however look at the graph. The increase of kinetic energy for the faster object, corresponds to a LOWER change of velocity, than for a body at rest.

Another way to think about it,is that a faster object spends LESS time in the field, moving through a distance x, so $\int \vec{F} \cdot dt$ is lower and thus a lower change of momentum.

This all changes when we consider relativistic speeds

$P = \frac{m_{0}v}{\sqrt{1-\frac{v^2}{c^2}}}$

Acceleration IS velocity dependant for a given net force.

$\endgroup$

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.