I know that: $$ K = \dfrac{1}{2}mv^2. $$ However, I was wondering if you can calculate the kinetic energy of an accelerating object and how could that be achieved?
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$\begingroup$ I don't see why you cannot use the same expression for kinetic energy in this case too. The only thing is- the kinetic energy is changing in time $\endgroup$– PaddyCommented Jul 18, 2020 at 18:50
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3 Answers
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Sure you can calculate the kinetic energy of an accelerating object. The acceleration is defined as change in velocity:
$$a=\frac{dv}{dt}$$
If you have some given acceleration $a(t)$ you can integrate this equation in order to get the velocity, which you can then pluck into the expression for the kinetic energy. In the simplest case of constant acceleration $a(t)=a_0$, the velocity is:
$$v(t)=a_0t+v_0$$
where $v_0$ is the velocity at $t=0$. The kinetic energy is then dependent on time:
$$E_\mathrm{kin}=\frac{m}{2}(a_0t+v_0)^2$$
answered Feb 3, 2017 at 19:48
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The kinetic energy of an accelerating object is simply $E = \frac{1}{2}mv^2$ for $v$ at that instance in time. The difference with an accelerating object is that that $v$ is constantly changing.
For a bit on context, Newton's second law tells us that an acceleration must be caused by a resultant force $F = ma$, when a force moves something a distance we have work $W = Fd$, this work is equivalent to the increase in energy of the moving object, not all of it is converted to kinetic energy, there may be heat, sound, gravitational potential etc. The point is that the Force causes the acceleration, which changes the velocity, which changes the kinetic energy, but we can also think of this as the force doing work on the object, which causes the change in kinetic energy.
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Sure. The formula remains same, except the kinetic energy becomes a function of time, since the object is accelerating: $$K(t) = \frac{1}{2} m \,v(t)^2 $$ If you have an explicit expression for the acceleration as a function of time, you could express the instantaneous velocity as $v(t) = \int_{0}^t a(t) \, dt$ and insert this into the first equation to yield $$K(t) = \frac{1}{2} m \,\left(\int_{0}^t a(t) \, dt\right)^2 $$ This is about as much as you can do without knowing what $a(t)$ is.
As a simple example, if the object is accelerating with a uniform acceleration $a(t) = a_0$, you could solve the integral for $v(t) = v_0 + a_0 t$ where $v_0$ is just some velocity at the start, giving $$K = \frac{1}{2} m \left (v_0 + a_0 t \right)^2. $$
