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

<p>The work done by a force is given by the integral of the force over the displacement. The force is given by:</p> <p>$F(x) = \alpha + \beta x^2$</p> <p>To find the work done when the object is displaced from $x = 0$ to $x = 1$, we compute:</p> <p>$W = \int_{0}^{1} (\alpha + \beta x^2) \, dx$</p> <p>Substituting $\alpha = 1 \, \text{N}$:</p> <p>$$ W = \int_{0}^{1} (1 + \beta x^2) \, dx = \int_{0}^{1} 1 \, dx + \int_{0}^{1} \beta x^2 \, dx $$</p> <p>Calculating each integral separately:</p> <p><p>$ \int_{0}^{1} 1 \, dx = [x]_{0}^{1} = 1 $</p></p> <p><p>$ \int_{0}^{1} \beta x^2 \, dx = \beta \left[ \frac{x^3}{3} \right]_{0}^{1} = \beta \left[\frac{1^3}{3} - \frac{0^3}{3}\right] = \frac{\beta}{3} $</p></p> <p>Thus, the total work done is:</p> <p>$W = 1 + \frac{\beta}{3}$</p> <p>We are given that the work done is 5 J, so:</p> <p>$1 + \frac{\beta}{3} = 5$</p> <p>Subtract 1 from both sides:</p> <p>$\frac{\beta}{3} = 4$</p> <p>Multiply both sides by 3 to solve for $\beta$:</p> <p>$\beta = 12 \, \text{N/m}^2$</p> <p>Therefore, the value of $\beta$ is $\boxed{12 \, \text{N/m}^2}$. Thus, Option C $12 \, \text{N/m}^2$ is the correct answer.</p>