Question

A particle of mass 0.01 kg is moving along the +ve x axis under influence of the force F(x) = -k(1/2x^2),where k = 0.01 Nm^2. At t=0, it is at x = 1m and velocity = 0.
Find the time at which it reaches x = 0.25

Answer 1

Answer on Question #46368, Physics, Mechanics | Kinematics | Dynamics

Question

A particle of mass $0.01 \, \text{kg}$ is moving along the +ve x axis under influence of the force $F(x) = -k(1/2x^2)$, where $k = 0.01 \, \text{Nm}^2$. At $t=0$, it is at $x = 1 \, \text{m}$ and velocity $= 0$.

Find the time at which it reaches $x = 0.25$.

Solution

According to Newton's second law the acceleration is produced when a force acts on a mass.

$$a = \frac{F}{m}$$

In our case, we have acceleration as a function of distance

$$a(x) = \frac{-k}{2m} x^2$$

The general definition:

$$a(x) = \frac{dv(x)}{dt} = \frac{dv(x)}{dx} \frac{dx}{dt} = \frac{dv(x)}{dx} v(x)$$

Recombine the differentials:

$$a(x) \, dx = v(x) \, dv(x)$$

Integrate both parts:

$$\int_{x_0}^{x} a(x) \, dx = \int_{v_0}^{v} v(x) \, dv(x) = \frac{v^2 - v_0^2}{2}$$

Solve for $v(x)$:

$$v(x) = \sqrt{v_0^2 + 2 \int_{x_0}^{x} a(x) \, dx}$$

Thus, in our case when $v_0 = 0$, $x_0 = 1$, $x = 0.25$

$$v(x) = \sqrt{2 \int_{1}^{0.25} \frac{-k}{2m} x^2 \, dx} = \sqrt{\frac{-k}{m} \int_{1}^{0.25} x^2 \, dx} = \sqrt{\frac{-k}{m} \left( \frac{x^3}{3} \right)_{1}^{0.25}}$$

So,

$$v(0.25) = \sqrt{\frac{-0.01}{0.01} \left( \frac{0.25^3}{3} - \frac{1}{3} \right)} = \sqrt{0.328125} = 0.573 \, \text{m/s}$$

Now, let's find $t(x)$. We will first find it's derivative:

$$\frac {d t (x)}{d x} = \frac {1}{v (x)}$$

Again, split the differentials and integrate:

$$\int_ {t _ {0}} ^ {t} d t = \int_ {x _ {0}} ^ {x} \frac {d x}{v (x)}$$

Evaluate the leftmost integral:

$$t = t _ {0} + \int_ {x _ {0}} ^ {x} \frac {d x}{\sqrt {v _ {0} ^ {2} + 2 \int_ {x _ {0}} ^ {x} a (x) d x}}$$

Thus,

$$t = \int_ {0} ^ {0. 2 5} \frac {d x}{0. 5 7 3} = \frac {0 . 2 5}{0 . 5 7 3} = 0. 4 3 6 s$$

Answer: $t = 0.436s$

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