Question

The work done in displacing a particle from $y=a$ to $y = 2a$ by a force $F = -\frac{K}{y^2}$ acting along y-axis is

• A. $5K/8a$
• B. $14K/8a^3$
• C. $-K/a^2$
• D. $-K/2a$

Hint:

Work done by a variable force is calculated by integrating the force over the path of displacement. For motion along a single axis, this simplifies to $W = \int_{x_1}^{x_2} F(x) dx$. Remember the power rule for integration: $\int x^n dx = \frac{x^{n+1}}{n+1} + C$.

Answer 1

The Correct Option is D

Work done by a variable force $F(y)$ in displacing a particle along the y-axis from $y_1$ to $y_2$ is given by the integral:
$W = \int_{y_1}^{y_2} F(y) dy$.
In this problem, the force is $F(y) = -\frac{K}{y^2}$, the initial position is $y_1=a$, and the final position is $y_2=2a$.
$W = \int_{a}^{2a} -\frac{K}{y^2} dy$.
We can take the constant $-K$ out of the integral:
$W = -K \int_{a}^{2a} \frac{1}{y^2} dy = -K \int_{a}^{2a} y^{-2} dy$.
Now, we evaluate the integral:
$W = -K \left[ \frac{y^{-1}}{-1} \right]_{a}^{2a} = -K \left[ -\frac{1}{y} \right]_{a}^{2a}$.
$W = K \left[ \frac{1}{y} \right]_{a}^{2a}$.
Now, apply the limits of integration:
$W = K \left( \frac{1}{2a} - \frac{1}{a} \right)$.
$W = K \left( \frac{1-2}{2a} \right) = K \left( -\frac{1}{2a} \right) = -\frac{K}{2a}$.