Question

Consider a force \[ F=Kx^3 \] which acts on a particle at rest. The work done by the force for displacement of \(2\,\text{m}\) is \((K=2\,\text{N m}^{-3})\):

• A. \(10\,\text{J}\)
• B. \(4\,\text{J}\)
• C. \(100\,\text{J}\)
• D. \(8\,\text{J}\)

Hint:

For variable forces, work done is obtained using integration: \[ W=\int F(x)\,dx. \] Always apply the proper limits of displacement.

Answer 1

The Correct Option is D

Step 1: Understand why we must integrate for a variable force.
When force changes with position, we cannot simply use $W = F \times d$ with a single value of $F$.
Instead, we divide the displacement into tiny elements $dx$, compute work for each, and sum them up:
\[ W = \int_0^{x_f} F\,dx \]

Step 2: Substitute the given force expression.
The force is $F = Kx^3$ with $K = 2\,\text{N\,m}^{-3}$ and displacement from $x = 0$ to $x = 2\,\text{m}$:
\[ W = \int_0^{2} 2x^3\,dx \]

Step 3: Pull the constant outside and apply the power rule.
\[ W = 2 \int_0^{2} x^3\,dx = 2 \left[\frac{x^4}{4}\right]_0^2 \]

Step 4: Evaluate the definite integral at the limits.
\[ W = 2 \times \frac{2^4 - 0^4}{4} = 2 \times \frac{16}{4} = 2 \times 4 = 8\,\text{J} \]

Step 5: Check the units.
$[K] = \text{N\,m}^{-3}$, $[x^3] = \text{m}^3$, so $[Kx^3] = \text{N}$. Then $[W] = \text{N} \cdot \text{m} = \text{J}$. Units are correct.

Step 6: State the answer.
\[ \boxed{8\,\text{J}} \]