Question:
A force \(F=4x\) is applied to move an object from \(x=0\) to \(x=2\,m\), then the work done is:
A force \(F=4x\) is applied to move an object from \(x=0\) to \(x=2\,m\), then the work done is:
Show Hint
For variable force problems:
\[
W=\int F\,dx
\]
Area under the force-position graph also represents work done.
Updated On: Jun 17, 2026
- \(8\,J\)
- \(16\,J\)
- \(4\,J\)
- \(32\,J\)
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The Correct Option is A
Solution and Explanation
Concept:
When force varies with position, work done is calculated using integration:
\[
W=\int_{x_1}^{x_2}F(x)\,dx
\]
Given:
\[
F=4x
\]
Limits:
\[
x=0 \text{ to } x=2
\]
Step 1: Set up the integral for work done. \[ W=\int_0^2 4x\,dx \]
Step 2: Integrate the expression. \[ W=4\int_0^2 x\,dx \] \[ =4\left[\frac{x^2}{2}\right]_0^2 \] \[ =2[x^2]_0^2 \] \[ =2(4-0) \] \[ =8\,J \] Hence, \[ \boxed{8\,J} \]
Step 1: Set up the integral for work done. \[ W=\int_0^2 4x\,dx \]
Step 2: Integrate the expression. \[ W=4\int_0^2 x\,dx \] \[ =4\left[\frac{x^2}{2}\right]_0^2 \] \[ =2[x^2]_0^2 \] \[ =2(4-0) \] \[ =8\,J \] Hence, \[ \boxed{8\,J} \]
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