Question 30: Work, Energy and Power & Circular Motion

Question

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

Accepted Answer

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

Answer

$15 \mathrm{~N} / \mathrm{m}^2$

Answer

$10 \mathrm{~N} / \mathrm{m}^2$

Answer

$12 \mathrm{~N} / \mathrm{m}^2$

Answer

$8 \mathrm{~N} / \mathrm{m}^2$

Work, Energy and Power & Circular Motion

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

Choose the correct answer: