A force F = α + βx^2 acts on an object in the x-direction. The work done by the force is 5J

Question

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

(1) \(15 \mathrm{~N} / \mathrm{m}^{2}\)

(2) \(10 \mathrm{~N} / \mathrm{m}^{2}\)

(3) \(12 \mathrm{~N} / \mathrm{m}^{2}\)

(4) \(8 \mathrm{~N} / \mathrm{m}^{2}\)

Answer 1

Correct option is : (3) \(12 \mathrm{~N} / \mathrm{m}^{2}\) 

\(F=\alpha+\beta x^{2}\) 

Work done \(=\int F d x\) 

\( 5=\int\left(\alpha+\beta x^{2}\right) d x\) 

\( 5=\alpha x+\left.\frac{\beta x^{3}}{3}\right|_{0} ^{1}\) 

\( 5=\alpha+\frac{\beta}{3}[\alpha=1]\) 

\( 4=\frac{\beta}{3} \Rightarrow \beta=12 \mathrm{~N} / \mathrm{m}^{2}\)