Concepts Of Physics MCQ Edition [Volume 1]PhysicsWork and Energy
A particle is acted upon by a force F = a + bx in the x -direction, where a and b are constants. Determine the work done by this force during a displacement from x = 0 to x = d .
Options
- A(a + bd )d
- B( 1 2 a + bd )d
- C( 1 2 a + 1 2 bd )d
- D(a + 1 2 bd )d
Correct answer
D. (a + 1 2 bd )d
Step-by-step solution
The work done by a variable force F during a displacement from x₁ to x₂ is given by W = _ x₁ ^ x₂ F , dx . Given F = a + bx , x₁ = 0 , and x₂ = d . Substituting the values into the work integral: W = ₀^ d (a + bx) , dx W = [ ax + 1 2 bx^2 ]₀^ d W = ad + 1 2 bd^2 W = (a + 1 2 bd )d