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 find the value of \( \beta \), we need to calculate the work done by the force \( F = \alpha + \beta x^2 \) as it displaces an object by 1 meter.

The work done by a force is given by the integral of the force over the displacement:

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

Given:

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

We now calculate the integral:

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

Substituting the limits of integration:

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

Equating it to the given work done:

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

Solve for \( \beta \):

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

Therefore, the correct value of \( \beta \) is 12 N/m².

This matches the correct answer from the given options.

Answer 2

The force acting on the object is \( F = \alpha + \beta x^2 \). The work done by a variable force is given by: \[ W = \int_{0}^{x} F \, dx \] Substituting \( F = \alpha + \beta x^2 \) into the equation: \[ W = \int_{0}^{x} (\alpha + \beta x^2) \, dx \] We can evaluate the integral: \[ W = \alpha x + \frac{\beta x^3}{3} \] Given that \( W = 5 \, {J} \) when \( x = 1 \, {m} \), we substitute these values: \[ 5 = \alpha \cdot 1 + \frac{\beta \cdot 1^3}{3} \] Since \( \alpha = 1 \, {N} \), this simplifies to: \[ 5 = 1 + \frac{\beta}{3} \] Solving for \( \beta \): \[ 5 - 1 = \frac{\beta}{3} \] \[ 4 = \frac{\beta}{3} \] \[ \beta = 12 \, {N/m}^2 \] Thus, the value of \( \beta \) is \( \boxed{12 \, {N/m}^2} \).