Question

A body of mass \(1\ \text{kg}\) starts moving from rest under the action of a force which varies with displacement as \(F=2x+5\) in newtons. The work done by this force to displace the body from \(x=0\) to \(x=2\ \text{m}\) is

• A. \(8\ \text{J}\)
• B. \(10\ \text{J}\)
• C. \(12\ \text{J}\)
• D. \(14\ \text{J}\)

Hint:

For variable force, do not use \(W=Fs\) directly. Use \(W=\int F\,dx\).

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The work done by a variable force is calculated by integrating the force over the distance of displacement.
Since the force depends on position \(x\), we use the integral formula for work.
Step 2: Key Formula or Approach:
Work done \(W = \int_{x_1}^{x_2} F(x) dx\).
Here \(F(x) = 2x + 5\), \(x_1 = 0\), and \(x_2 = 2\).
Step 3: Detailed Explanation:

Set up the integral for work:
\[ W = \int_{0}^{2} (2x + 5) dx \]

Integrate the expression term by term:
\[ W = \left[ 2 \cdot \frac{x^2}{2} + 5x \right]_0^2 \]
\[ W = [ x^2 + 5x ]_0^2 \]

Evaluate at the upper and lower limits:
\[ W = (2^2 + 5(2)) - (0^2 + 5(0)) \]
\[ W = (4 + 10) - 0 \]
\[ W = 14 \text{ Joules} \]

Step 4: Final Answer:
The total work done by the variable force to move the body from \(x = 0\) to \(x = 2\) is 14 J.