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 N then $\beta$ will be

• A. 15 N/m²
• B. 12 N/m²
• C. 8 N/m²
• D. 10 N/m²

Answer 1

Answer: (B)

12 N/m²

Answer 2

The work done by a variable force $F(x)$ acting on an object in displacing it from $x_1$ to $x_2$ is given by the integral:

$W = \int_{x_1}^{x_2} F(x) dx$

Given:

• Force $F = \alpha + \beta x^2$
• Work done $W = 5$ J
• Displacement is 1 m. Assuming the object starts from $x_1 = 0$ m, it moves to $x_2 = 1$ m.
• Constant $\alpha = 1$ N

Substitute the given force function and limits into the work integral:

$W = \int_{0}^{1} (\alpha + \beta x^2) dx$

Evaluate the definite integral:

$W = \left[ \alpha x + \beta \frac{x^3}{3} \right]_{0}^{1}$

$W = \left( \alpha (1) + \beta \frac{(1)^3}{3} \right) - \left( \alpha (0) + \beta \frac{(0)^3}{3} \right)$

$W = \left( \alpha + \frac{\beta}{3} \right) - (0)$

$W = \alpha + \frac{\beta}{3}$

Now, substitute the given values for $W$ and $\alpha$:

$5 = 1 + \frac{\beta}{3}$

Solve for $\beta$:

$5 - 1 = \frac{\beta}{3}$

$4 = \frac{\beta}{3}$

$\beta = 4 \times 3$

$\beta = 12$ N/m²

The value of $\beta$ is 12 N/m².