Question

A 1 kg box placed at the origin starts sliding along x-axis under the action of a force \(\vec{F} = m \vec{a}\). Its acceleration as a function of x is given by \(a(x) = \beta x\) where \(\beta = 5 \, \text{s}^{-2}\). Find the work done by \(\vec{F}\) in moving the box from \(x = 2 \, \text{cm}\) to \(x = 5 \, \text{cm}\).

• A. \(52.5 \times 10^{-4} \, \text{J}\)
• B. \(105.5 \times 10^{-4} \, \text{J}\)
• C. \(17.0 \times 10^{-4} \, \text{J}\)
• D. \(34.0 \times 10^{-4} \, \text{J}\)

Hint:

For variable acceleration along x-axis, work done by force can be calculated using \(W = \int F dx = m \int a(x) dx\).

Answer 1

The Correct Option is A

Step 1: Go through the work-energy theorem instead of the force integral directly.
Since $a = \dfrac{dv}{dt} = v\dfrac{dv}{dx}$, we can write $v\,dv = a(x)\,dx = \beta x\,dx$. Integrating both sides tracks kinetic energy directly, and by the work-energy theorem, the change in kinetic energy across a displacement equals the work done by the force causing it.

Step 2: Integrate to get the change in kinetic energy per unit mass.
\[ \int v\,dv = \beta\int x\,dx \implies \Delta\left(\frac{v^2}{2}\right) = \frac{\beta}{2}\left(x_2^2 - x_1^2\right) \]
Multiplying through by the mass $m$ gives the work done directly, $W = \Delta KE = \dfrac{m\beta}{2}(x_2^2 - x_1^2)$, so this route lands on the same integral as a direct force calculation but arrives there through kinetic energy.

Step 3: Plug in $m = 1$ kg, $\beta = 5\ \text{s}^{-2}$, $x_1 = 0.02$ m, $x_2 = 0.05$ m.
\[ W = \frac{1 \times 5}{2}\left(0.05^2 - 0.02^2\right) = 2.5\left(0.0025 - 0.0004\right) = 2.5 \times 0.0021 \]

Step 4: Compute.
\[ W = 0.00525\ \text{J} = 52.5\times10^{-4}\ \text{J} \]

Step 5: Conclusion.
\[ \boxed{52.5\times10^{-4}\ \text{J}} \]