Question

A position dependent force, F = (7 - 2x + 3x$^2$) N acts on a small body of mass 2 kg and displaces it from x = 0 to x = 5 m. The work done in joule is

• A. 135
• B. 270
• C. 35
• D. 70

Answer 1

The Correct Option is A

To calculate the work done by a variable force \( F(x) = 7 - 2x + 3x^2 \) in displacing an object from position \( x = 0 \) to \( x = 5 \), we use the principle of work done by a force, which is given by the integral of the force over the displacement:

W = \int_{x_0}^{x_1} F(x) \, dx

Substituting for \( F(x) \) and the limits of integration, the work done is:

W = \int_{0}^{5} (7 - 2x + 3x^2) \, dx

Let us evaluate this integral step-by-step:

1. Integrate each term individually:
   • \int 7 \, dx = 7x
   • \int -2x \, dx = -x^2 (since \(-2 \times \frac{x^2}{2} = -x^2\))
   • \int 3x^2 \, dx = x^3 (since \(3 \times \frac{x^3}{3} = x^3\))
2. Combine the results:

   W = \left[ 7x - x^2 + x^3 \right]_{0}^{5}

3. Evaluate the expression at the limits:

   Substitute \( x = 5 \):

   W(5) = 7(5) - (5)^2 + (5)^3 = 35 - 25 + 125 = 135

   Substitute \( x = 0 \):

   W(0) = 7(0) - (0)^2 + (0)^3 = 0

4. Calculate the net work done:

   The work done \( W \) is the difference between the values at the limits:

   W = W(5) - W(0) = 135 - 0 = 135 \, \text{Joules}

Hence, the work done is 135 Joules.