Question

A particle moves along X-axis from $x=0$ to $x=5$ cm under the influence of force $F=(7-2x+3x²)$ N. The work done is:

• A. 70 J
• B. 270 J
• C. 35 J
• D. 135 J

Hint:

Work done = $\int Fdx$.

Answer 1

The Correct Option is D

To solve this problem, we need to calculate the work done by the force \(F(x) = (7 - 2x + 3x^2)\) N as the particle moves from \(x = 0\) to \(x = 5\) cm along the X-axis.

The work done by a force on an object moving along a straight path is given by the integral of the force over the path of motion. Here, work done \(W\) is calculated as:

\(W = \int_{x=0}^{x=5} F(x) \, dx\)

Substitute \(F(x)\) into the integral:

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

Now, calculate the integral:

1. Integrate term by term:
   • The integral of \(7\) is \(7x\).
   • The integral of \(-2x\) is \(-x^2\).
   • The integral of \(3x^2\) is \(x^3\).
2. Substitute the integration limits \(0\) and \(5\) into the resulting expression:

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

Calculate:

• At \(x = 5\), \(7(5) - (5)^2 + (5)^3 = 35 - 25 + 125 = 135\).
• At \(x = 0\), \(7(0) - (0)^2 + (0)^3 = 0\).

Thus, the work done is:

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

Hence, the correct answer is 135 J, which matches the given correct answer.