I've never been so confused (Application of Integral Calculus)

Question

Here's a problem on Application of Integral calculus to find the work done in moving a particle. I was able to 'reach' the 'right answer'. But I'm totally confused and utterly dissatisfied with the way I did it. It was like I found the way to end up at given answer and not the other way round.

The Question:
A particle of mass $m$ starts from rest at time $t=0$ and is moved along the $x$-axis with constant acceleration $a$ from $x=0$ to $x=h$ against a variable force of magnitude $F(t)=t^2$. Find the work done.

The answer to this question is given as $$ \frac{4h\sqrt{3mh}}{3} $$

My work:
The way to reach the given answer is simple. Let $$ \begin{aligned} F(t)&=t^2 \\ &=ma \\ \implies a=\frac{t^2}{m} \end{aligned} $$

Integrating this from $0$ to $t$, we get the velocity as a function of $t$ and doing it again, we get displacement as a function of t. $$ v(t)=\frac{t^3}{3m} \\ x(t)=\frac{t^4}{12m} $$

Time at which the the particle is at $x=h$ would be $$ t_h=\sqrt[4]{12mh} $$ Work done on the particle is the change in its kinetic energy between $t=0$ and $t=t_h$ $$ \begin{aligned} W &= \frac{1}{2}m\big[v(t_h)\big]^2-\underbrace{\frac{1}{2}m\big[v(0)\big]^2}_{=0} \\ &= \frac{1}{2}m \left\{\frac{t_h^3}{3m}\right\}^2 \\ &= \frac{1}{2}m \left\{\frac{(\sqrt[4]{12mh})^3}{3m}\right\}^2 \\ &= \frac{4h\sqrt{3mh}}{3} \end{aligned} $$

My Confusions:
The first equation that I've used: $$ a=\frac{t^2}{m} $$ goes against the problem statement. In the problem, $a$ is constant. But here I've taken it as a function of time and consequently, variable (since mass is constant).

My line of thought goes like this: If the particle is moving at a constant acceleration $a$ against the force $F(t)=t^2$, then there must be another force, say $F_1$, which should also be variable with time, that is overwhelming $F(t)$ to make the particle move with that constant acceleration. Right? $$ ma=F_1-F(t)=ma_1-t^2 $$ where $a_1$ should be variable with time.

If we continue with this, then we won't end up with the answer that is given. I'm really, really confused. Can someone help?

Answer 1

You're right to be suspicious; the question is just all sorts of wrong. The question says

A particle of mass $m$ starts from rest at time $t=0$ and is moved along the $x$-axis with constant acceleration $a$ from $x=0$... blablablablablablabla

Once I read this first part of the sentence, without even thinking, I immediately translate this into an initial value problem for a second order ODE:

\begin{align} \begin{cases} x''(\cdot)&=a\\ x'(0)&=0\\ x(0)&=0 \end{cases} \end{align} The solution is immediately given as $x(t)=\frac{1}{2}at^2$, and so it takes a total of $t_h=\sqrt{\frac{2h}{a}}$ units of time to travel a displacement of $h$. It doesn't matter how many forces or what kinds of forces act on the particle. If the particle has a constant acceleration of $a$, then this is the trajectory of the particle. That's it.

Next, the net work done along the path from $x=0$ to $x=h$ is simply the change in kinetic energy (as you've correctly identified), and in this case, it is \begin{align} \frac{m}{2}[x'(t_h)]^2-\frac{m}{2}[x'(0)]^2=\frac{m}{2}[at_h]^2-\frac{m}{2}(0)^2=mah. \end{align} If you wish to calculate only the work done by the force $F$, then it is \begin{align} \text{work done by $F$ on the particle}&=\int_0^{t_h}F(t)\cdot x'(t)\,dt=\int_0^{t_h}t^2\cdot at\,dt=\frac{at_h^4}{4}, \end{align} which again is not the "intended" answer. In the current form of how the question is phrased, you're right to introduce a new force $F_1$, but even the work done by $F_1$ is not equal to the "intended" answer. i.e none of the forces $F, F_1, F_1-F$ have work equal to the "intended" answer. This just goes to show how poorly the question is worded.

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If one wants to obtain the "intended" answer, here's how the question should have been phrased:

Consider a particle of mass $m$ which undergoes motion along the $x$-axis; let $x(t)$ denote the position of the particle at time $t$. Suppose that the particle is under the influence of a force $F$, such that $F(x(t))=kt^2$ for some constant $k>0$ (to ensure that $F$ has the right units) and that the particle starts off at rest at the origin. Find the work done by the force $F$ on the particle as it is displaced from $0$ to some final displacement $h$.

I'm pretty sure here the answer works out to be $\frac{4}{\sqrt{3}}\sqrt{mh^3k}$ (the same as what you have, up to the factor of $k$).