Question

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

• A. 10 J
• B. 4 J
• C. 100 J
• D. 8 J

Hint:

For a variable force $F(x)$, the work done over a displacement is the integral $\int F(x) \, dx$. Ensure the limits of integration correspond to the initial and final positions of the displacement.

Answer 1

The Correct Option is D

The force acting on the particle is $F = K x^3$, where $K = 2 \, \text{N m}^{-3}$, and the particle is displaced from $x = 0$ (since it starts at rest, we assume the initial position is $x = 0$) to $x = 2 \, \text{m}$. We need to find the work done by the force over this displacement. Work done by a variable force is given by: \[ W = \int_{x_1}^{x_2} F \, dx \] Substitute $F = K x^3$ and $K = 2 \, \text{N m}^{-3}$, with limits from $x = 0$ to $x = 2$: \[ W = \int_0^2 (2 x^3) \, dx = 2 \int_0^2 x^3 \, dx \] Integrate: \[ \begin{align} \int x^3 \, dx = \frac{x^4}{4}, \quad \text{so} \quad \int_0^2 x^3 \, dx = \left[ \frac{x^4}{4} \right]_0^2 = \frac{2^4}{4} - \frac{0^4}{4} = \frac{16}{4} = 4 \] \[ W = 2 \times 4 = 8 \, \text{J} \] The work done by the force is 8 J, which matches option (4). Thus, the correct answer is (4).