A force $F=\alpha+\beta x^{2}$ acts on an object in the $x$-direction. The work done by the force is

Question

A force $F=\alpha+\beta x^{2}$ 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 $\alpha=1 \mathrm{~N}$ then $\beta$ will be\

Professor PC Thomas Classes · Answer

(d) : Applied force, $F=\alpha+\beta x^{2}$ Displacement $=1 \mathrm{~m}$\ World done by force, $W=5 \mathrm{~J}$\ $\because \quad W=\int_{0}^{x} F d x \Rightarrow 5 \mathrm{~J}=\int_{0}^{1}\left(\alpha+\beta x^{2}ight) d x$\ $\Rightarrow\left|\alpha x+\frac{\beta x^{3}}{3}ight|_{0}^{1}=5 \mathrm{~J}$\ $\Rightarrow 1+\frac{\beta}{3}=5 \quad(\because \alpha=1 \mathrm{~N})$\ $\Rightarrow \beta=12 \mathrm{~N} / \mathrm{m}^{2}$\