Work done by a variable force

Q. Work done by a variable force

Question.

When a particle is located a distance x meters from the origin, a force of cos(πx/3) newton acts on it. How much work is done in moving the particle from x = 1 to x = 2 ? Interpret your answer by considering the work done from x = 1 to x = 1.5 and from x = 1.5 to x = 2.

Answer. 

Work done 

= integral (x1 to x2) cos (πx/3) dx 

1)

= (3/π) sin (πx/3) [ x = 1 to x = 2 ]

= (3/π) [sin (2π/3) - sin (π/3)]
= (3/π) (1/2 - 1/2) = 0

2)

= (3/π) sin (πx/3) [ x = 1 to x = 1.5 ]

= (3/π) [sin(π/2) - sin(π/3)]

= (3/π) (1 - 1/2)

= 3/(2π)

3)

= (3/π) sin (πx/3) [ x = 1.5 to x = 2.0 ]

= (3/π) [sin(2π/3) - sin(π/2)]

= (3/π) (1/2 - 1)

= - 3/(2π)

Interpretation:

This is the case of a variable force. The direction of force changes at x = 1.5. So work done is positive from x = 1 to x = 1.5. Same amount of negative work is done as the force is negative for x = 1.5 to x = 2. Hence, total work done for x = 1 to x = 2 is zero.