Question

A force F = $\alpha$ + $\beta$x² 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 displaced 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 is displaced from $x_1 = 0$ m to $x_2 = 1$ m. Constant $\alpha = 1$ N

Substitute the given values into the work done formula: $5 = \int_{0}^{1} (1 + \beta x^2) dx$

Now, evaluate the definite integral: $5 = \left[ x + \frac{\beta x^3}{3} \right]_{0}^{1}$

Apply the limits of integration: $5 = \left( 1 + \frac{\beta (1)^3}{3} \right) - \left( 0 + \frac{\beta (0)^3}{3} \right)$ $5 = \left( 1 + \frac{\beta}{3} \right) - (0)$ $5 = 1 + \frac{\beta}{3}$

Now, 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.