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Concepts Of Physics MCQ Edition [Volume 1]
Exam
Concepts Of Physics MCQ Edition [Volume 1]
Subject
Physics
Chapter
Work and Energy

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Work and Energy

A particle is acted upon by a force F = a + bx in the x -direction,...

A particle is acted upon by a force F=a+bxF = a + bx in the xx-direction, where aa and bb are constants. Determine the work done by this force during a displacement from x=0x = 0 to x=dx = d.

Options

  1. A\left(a + bd\right)d
  2. B\left(\dfrac{1}{2}a + bd\right)d
  3. C\left(\dfrac{1}{2}a + \dfrac{1}{2}bd\right)d
  4. D\left(a + \dfrac{1}{2}bd\right)d

Correct answer

D. \left(a + \dfrac{1}{2}bd\right)d

Step-by-step solution

The work done by a variable force FF during a displacement from x1x_1 to x2x_2 is given by W=∫x1x2F dxW = \int_{x_1}^{x_2} F \, dx.

Given F=a+bxF = a + bx, x1=0x_1 = 0, and x2=dx_2 = d.

Substituting the values into the work integral:

W=∫0d(a+bx) dxW = \int_{0}^{d} (a + bx) \, dx

W=[ax+12bx2]0dW = \left[ ax + \dfrac{1}{2}bx^2 \right]_{0}^{d}

W=ad+12bd2W = ad + \dfrac{1}{2}bd^2

$W = \left(a + \dfrac{1}{2}bd\right)d
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