Physics · Work, Energy And Power · NEET
W = F d cos theta only works when the force stays constant over the whole path. If the force value changes as the particle moves (for example F depends on position x), then there is no single F to plug in. The fix is to break the path into tiny pieces so small that F is almost constant over each piece, find work F(x) dx on each piece, and add them all. That sum is the integral W = integral of F dx.
The limits are the starting position and the ending position of the particle along the direction of the force. If the particle moves from x = a to x = b, then W = integral from a to b of F(x) dx. Always put the initial position as the lower limit and the final position as the upper limit. For the NEET 2019 problem F = 20 + 10y, the particle goes from y = 0 to y = 1, so the limits are 0 and 1.
To find work, integrate with respect to position (dx or dy), not time. Work is force times displacement, so the variable must be the displacement variable. Only if you want power do you bring in time. If force is given as F(x), write W = integral of F(x) dx directly.
The integral W = integral of F dx is exactly the area between the F-x curve and the x-axis. So integration (this page) and the graph method (area method) give the same answer. Use integration when F is given as a formula; use the area method when F is given as a graph or straight-line segments.
The method is identical, just replace x with y. The particle moves along y, so W = integral of F(y) dy between the y limits. The letter (x or y) is only the name of the displacement direction; the integration steps are the same.
A force F = 20 + 10y acts on a particle in the y-direction, where F is in newton and y in metre. The work done by this force to move the particle from y = 0 to y = 1 m is
Try the real previous-year questions from this chapter — each with the answer and a full solution.
W = integral of F(x) dx taken from the initial position to the final position. In words, add up F(x) dx over every tiny step of the path.
Work is always a scalar. Even though force is a vector, the dot product F dot dx gives a scalar, so the total work is just a number with unit joule.
Use integration when F is given as a formula in terms of position. Use the area under the F-x graph when force is given as a graph or as straight-line segments where area is easy to compute. Both give the same value.
Yes. If the force opposes the displacement over some part, F(x) is negative there, and that region contributes negative work. Springs and friction often give negative work regions.
Yes. The spring force F = -kx is a variable force, and its work W = integral of -kx dx = -(1/2)k x^2 is found by exactly this integration method.
