I have learned that $$\mathbf{F} = m \mathbf{a}$$ where both the force $\mathbf{F}$ and acceleration $\mathbf{a}$ are vectors. This makes sense, since both force and acceleration have a direction.

On the other hand, the kinetic energy $$K = \frac12 mv^2$$ looks completely different. It doesn't seem to depend on the direction. How are these two concepts related?


When a force is applied over a certain distance, that force does mechanical work, $W$. If the force is constant $F$ and the object it is exerted on is moved by a distance $\Delta x$, then $W=F\Delta x$. If the force is not constant but a function of the position, this turns into an integral: $$W = \int_{x_1}^{x_2}F(x)\,\mathrm d x.$$ If you don't know calculus yet, just ignore this.

Note, that is is not important how long (in time) the force is exerted. E.g. a cup on a table will feel the constant force due to gravity but is won't move (because the table is pushing it upwards with an equal but opposite force) so there is no work done on that cup, meaning that its energy content won't change.

Work is basically just energy change. Depending on how the work is applied, it will increase (or decrease) a specific kind of energy. If the work leads to a change in the (absolute) velocity, it will modify the kinetic energy.

E.g. if a car accelerates from standstill with constant acceleration $a$ (i.e. the engine will exert a constant forward force on the car), its velocity increases linearly in time, $v(t)=vt$ and its position quadratically, $x(t)=\frac{1}{2}at^2$. After a time $t_1$, it went over a distance $x_1=x(t_1)=\frac{1}{2}at_1^2$, the work done by the engine will be $Fx_1 = max_1= \frac{1}{2}ma^2t_1^2$. At time $t_1$, the velocity of the car is $v_1=v(t_1)=at_1$, so we can write the work done by the engine as $\frac{1}{2}mv_1^2$. This is exactly the amount of kinetic energy gained by the car. So the work done by the engine was used to increase the car's kinetic energy.


It is really a problem of definition; the kinetic energy defines a useful quantity which by definition is a scalar not a vector. You don't actually strictly speaking apply a kinetic energy to a body. A body carries a kinetic energy by the mere virtue of its speed and there is a difference between speed and velocity. This quantity Kinetic energy can be used in equations such as the mechanical energy conservation or the work-energy theorem etc. Velocity refers to a vector including information about magnitude and direction. Speed does not! Speed only carries info about magnitude and note here that the v in the kinetic energy formula refers only to speed.


As you will find out, each equation is applicable to different type of problems, not related to each other. So I am going to assume you are just "curious" about the relationship (if any), between force (F)and kinetic energy (E).
To "come up" with the relationship, all you have to do is divide one equation by the other $$E/F = (1/2mv^2)/(ma)$$ Since the mass term appears in both numerator and denominator, it cancels, leaving $v^2$/2a, and since a = v/t, the relationship formula becomes: $$E/F = vt/2$$


MD of MKForce

RE the formula E = 1/2 m v squared and F = mass x acceleration

The E = formula is related to velocity and F = ma is related to acceleration.

So in other words the rate of change of Force is proportion to the rate of change of acceleration multiplied by the scaler mass.

If we divide the two formulas we arrive at :

E/F= the rate of change of energy in relation to force, = 1/2 m v sq / m a The masses cancel out leaving

E/F = 1/2 v squared / acceleration

acceleration = v/t the rate of change of velocity with respect to time

E/F = 1/2 v squared/a = 1/2 v sq / v/t

So Force = E / 1/2 v t = 2E / v t So Force is the twice rate of change of Energy with respect to the velocity x time

But as we know Albert said E= m c squared.

We put this into the equation

Force = 2 m c squared /vt

So if m and c are constant the force is the inverse of the velocity x time (1 / vt) scaled up by the mass x the speed of light squared. Or the 2mc sq scaled down by the velocity x time. At a velocity of c squared Force = m t

Force = c sq/v x 2 m t

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The full equation goes as follows:

D = -(M/2K) ln ( (T - Mg - Kv^2)/(T - Mg) )

where: D = distance traveled, M = mass of the subject, g = acceleration due to gravity, v = initial speed/velocity, T = thrust, and K = drag factor (in units mass/distance).

If K = 0, then l'Hopital gives (remember to take the constant -M/2 outside the limit):

lim(K->0) D(K) = -(M/2) lim ( (T - Mg)/(T - Mg - Kv^2) ) (-v^2/(T - Mg) ) = (Mv^2/2)/(T - Mg)

That is, D = kinetic energy / net force (thrust minus gravity) when there's no drag.

  • 2
    $\begingroup$ Please render using LaTex. This is barely readable. $\endgroup$ – Gert Dec 23 '15 at 2:49
  • $\begingroup$ WTH the OP is in 8th standard $\endgroup$ – Dove Apr 5 '17 at 12:12

protected by AccidentalFourierTransform Feb 9 '18 at 18:05

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