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 \((K=2\,\text{N m}^{-3})\):
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 \((K=2\,\text{N m}^{-3})\):
Show Hint
For variable forces, work done is obtained using integration:
\[
W=\int F(x)\,dx.
\]
Always apply the proper limits of displacement.
Updated On: Jun 24, 2026
- \(10\,\text{J}\)
- \(4\,\text{J}\)
- \(100\,\text{J}\)
- \(8\,\text{J}\)
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Write the expression for work done by a variable force.
For a variable force, \[ W=\int F\,dx \] Given, \[ F=Kx^3 \] Hence, \[ W=\int_0^2 Kx^3\,dx \]
Step 2: Substitute the value of \(K\).
Since \[ K=2\,\text{N m}^{-3}, \] we get \[ W=\int_0^2 2x^3\,dx \] \[ W=2\int_0^2 x^3\,dx \]
Step 3: Integrate.
\[ W=2\left[\frac{x^4}{4}\right]_0^2 \] \[ W=2\left(\frac{2^4}{4}\right) \] \[ W=2\left(\frac{16}{4}\right) \] \[ W=2(4) \] \[ W=8\,\text{J} \]
Step 4: Final conclusion.
Hence, the work done is \[ \boxed{8\,\text{J}} \]
For a variable force, \[ W=\int F\,dx \] Given, \[ F=Kx^3 \] Hence, \[ W=\int_0^2 Kx^3\,dx \]
Step 2: Substitute the value of \(K\).
Since \[ K=2\,\text{N m}^{-3}, \] we get \[ W=\int_0^2 2x^3\,dx \] \[ W=2\int_0^2 x^3\,dx \]
Step 3: Integrate.
\[ W=2\left[\frac{x^4}{4}\right]_0^2 \] \[ W=2\left(\frac{2^4}{4}\right) \] \[ W=2\left(\frac{16}{4}\right) \] \[ W=2(4) \] \[ W=8\,\text{J} \]
Step 4: Final conclusion.
Hence, the work done is \[ \boxed{8\,\text{J}} \]
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