Q. A force $F=\alpha+\beta x^2$ acts on an object in the $x$-direction. The work done by the force is $5\,\text{J}$ when the object is displaced by $1\,\text{m}$. If the constant $\alpha=1\,\text{N}$ then $\beta$ will be
A) $15\,\text{N/m}^2$
B) $12\,\text{N/m}^2$
C) $8\,\text{N/m}^2$
D) $10\,\text{N/m}^2$
Explanation
[Solution by Dr. Abubacker] $W=\int_0^1(\alpha+\beta x^2)dx=\alpha+\frac\beta3=1+\frac\beta3=5\Rightarrow\beta=12\ \text{N/m}^2$. Answer (2).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.