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. 10 N/m²
• C. 12 N/m²
• D. 8 N/m²

Hint:

When dealing with variable forces, the work done is calculated by integrating the force over the displacement. In this case, the force was given as \( F = \alpha + \beta x^2 \), and the integral was used to find the work done.

Answer 1

The Correct Option is C

To determine the value of \( \beta \), we must compute the work performed by the force \( F = \alpha + \beta x^2 \) over a displacement of 1 meter.

Work done by a force is defined as the integral of the force with respect to displacement:

\(W = \int F \, dx = \int (\alpha + \beta x^2) \, dx\)

The following are given:

• \( \alpha = 1 \, \text{N} \)
• The total work done \( W = 5 \, \text{J} \)
• The displacement occurs from \( x = 0 \) to \( x = 1 \, \text{m} \)

We now evaluate the integral:

\(W = \int_0^1 (1 + \beta x^2) \, dx = \left[ x + \frac{\beta x^3}{3} \right]_0^1\)

Applying the limits of integration:

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

Equating this result to the given work done:

\(1 + \frac{\beta}{3} = 5\)

Solving for \( \beta \):

• Subtract 1 from both sides: 
  \(\frac{\beta}{3} = 4\)
• Multiply both sides by 3: 
  \(\beta = 12\) N/m²

Consequently, the determined value for \( \beta \) is 12 N/m².

This value corresponds to the correct answer among the provided options.