Question:
A body of mass \(1\ \text{kg}\) starts moving from rest under the action of a force which varies with displacement as \(F=2x+5\) in newtons. The work done by this force to displace the body from \(x=0\) to \(x=2\ \text{m}\) is
A body of mass \(1\ \text{kg}\) starts moving from rest under the action of a force which varies with displacement as \(F=2x+5\) in newtons. The work done by this force to displace the body from \(x=0\) to \(x=2\ \text{m}\) is
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For variable force, do not use \(W=Fs\) directly. Use \(W=\int F\,dx\).
Updated On: May 7, 2026
- \(8\ \text{J}\)
- \(10\ \text{J}\)
- \(12\ \text{J}\)
- \(14\ \text{J}\)
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The Correct Option is D
Solution and Explanation
Concept:
When force varies with displacement, work done is:
\[
W=\int_{x_1}^{x_2}F\,dx
\]
Step 1: Given force: \[ F=2x+5 \] Displacement is from: \[ x=0 \quad \text{to} \quad x=2 \]
Step 2: Write work done: \[ W=\int_0^2(2x+5)\,dx \]
Step 3: Integrate. \[ W=\left[x^2+5x\right]_0^2 \]
Step 4: Apply limits. \[ W=(2^2+5(2))-(0^2+5(0)) \] \[ W=4+10 \] \[ W=14\ \text{J} \] Therefore, \[ \boxed{14\ \text{J}} \]
Step 1: Given force: \[ F=2x+5 \] Displacement is from: \[ x=0 \quad \text{to} \quad x=2 \]
Step 2: Write work done: \[ W=\int_0^2(2x+5)\,dx \]
Step 3: Integrate. \[ W=\left[x^2+5x\right]_0^2 \]
Step 4: Apply limits. \[ W=(2^2+5(2))-(0^2+5(0)) \] \[ W=4+10 \] \[ W=14\ \text{J} \] Therefore, \[ \boxed{14\ \text{J}} \]
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