Question

The force acting on a particle in the \(x\)-direction is \[ (3+2x)\ \text{N}, \] where \(x\) is the displacement of the particle in metre. The work done in displacing the particle from \[ x=1.5\ \text{m} \text{ to } x=3.5\ \text{m} \] is

• A. \(24\,\text{J}\)
• B. \(32\,\text{J}\)
• C. \(16\,\text{J}\)
• D. \(8\,\text{J}\)

Hint:

When force varies with displacement, \[ \boxed{W=\int_{x_1}^{x_2}F(x)\,dx.} \] The work done equals the area under the force--displacement graph.

Answer 1

The Correct Option is C

Step 1: Use the definition of work done by a variable force. For a variable force, \[ W=\int_{x_1}^{x_2}F(x)\,dx. \] Here, \[ F(x)=3+2x, \] with \[ x_1=1.5,\qquad x_2=3.5. \] Hence, \[ W=\int_{1.5}^{3.5}(3+2x)\,dx. \]

Step 2: Evaluate the integral. Integrating, \[ W=\left(3x+x^2\right)_{1.5}^{3.5}. \] At \[ x=3.5, \] \[ 3(3.5)+(3.5)^2 =10.5+12.25 =22.75. \] At \[ x=1.5, \] \[ 3(1.5)+(1.5)^2 =4.5+2.25 =6.75. \] Therefore, \[ W=22.75-6.75=16\,\text{J}. \] Hence, \[ \boxed{W=16\,\text{J}.} \] Therefore, the correct option is \(\boxed{(C)}\).