Work Energy Power

JEE Main level, single correct, medium

6 of 183

Question 6 of 183

A force F=α+βx2\mathrm{F}=\alpha+\beta \mathrm{x}^2 acts on an object in the x -direction. The work done by the force is 5 J when the object is displaced by 1 m . If the constant α=1 N\alpha=1 \mathrm{~N} then β\beta will be

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Answer: (B) 12 N/m212 \mathrm{~N} / \mathrm{m}^2

F=α+βx2F=\alpha+\beta x^2
Work done ∫dw=∫F⋅dx\int d w=\int F \cdot d x
$\begin{aligned}
& \Rightarrow \quad \Delta W=\int F \cdot d x=\int\left(\alpha+\beta x^2\right) d x \\
& \Rightarrow \quad \Delta W=\left|\alpha x+\frac{\beta x^3}{3}\right|_0^1=\alpha+\frac{\beta}{3}=5
\end{aligned}$
Given α=1\alpha=1
So, β3=4\frac{\beta}{3}=4
⇒β=12 N/m2\Rightarrow \beta=12 \mathrm{~N} / \mathrm{m}^2

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