Physics problem regarding a sliding box and variable retardation

Question

I am currently working on a physics problem that asks to solve for the velocity at a certain distance when a box is slid with an initial velocity of 10 m/s. The problem also gives a variable force that is retarding the motion of the box. This equation is $f(x) = m(-15+.28x^3)$ where m is mass and x is distance traveled.

Given a distance of 4, I wanted to relate that Force is F=MA and that the $(-15+.28x^3)$ would be acceleration and I could use a kinematics equation; $V_f^2=V_i^2+2ax$. This would give me a final velocity of ~7m/s. Another part of the problem asks how far the box will go before it stops. This is where I am stuck on. I would appreciate any help on this problem.

Answer 1

Rewrite Newton's second law as $$ \frac{1}{m}f(x)=\frac{dv}{dt}=\frac{dx}{dt}\frac{dv}{dx} =v\frac{dv}{dx}. $$ This is an easy to solve differential equation.

Answer 2

Begin with Newton's Law:

$$\vec{F}_\text{net} = \frac{d \vec{p}}{d t} = m \frac{d \vec{v}}{d t}$$

In this problem, the motion is in one dimension, so we can simplify $\vec{F}_\text{net} = f \vec{e}_x$ and $\vec{v} = v \vec{e}_x$. Then dotting the equation with $\vec{e}_x$ gives

$$f = m \frac{d v}{d t}$$

Here, we are given $f$ as a function of $x$ and desire $v$ as a function of $x$. Therefore, we should use the chain rule to write

$$\frac{d v}{d t} = \frac{d v}{d x} \frac{d x}{d t} = v \frac{d v}{d x}$$

so that we are solving

$$f = m v \frac{d v}{d x}$$

Here, we take $f = m (-C_1 + C_2 x^3)$, so that

$$-C_1 + C_2 x^3 = v \frac{d v}{d x}$$

with initial condition $v (x = 0) = v_0$. This ODE is separable:

$$\int_0^x (-C_1 + C_2 x^3) d x = \int_0^{v(x)} v d v$$

$$\left[ -C_1 x + \frac{1}{4} C_2 x^4 \right]_0^x = \left[ \frac{1}{2} v^2 \right]_0^{v(x)}$$

$$-C_1 x + \frac{1}{4} C_2 x^4 = \frac{1}{2} (v(x)^2 - v_0^2)$$

$$v(x)^2 = v_0^2 - 2 C_1 x + \frac{1}{2} C_2 x^4$$

Note that because of the quadratic term, two velocities are possible for a given position: one where the particle is moving to the left and the other where the particle is moving to the right. You will need to determine which one is true.