Q.While decreasing linearly from 5 N to 3 N, a force displaces an object from 3 m to 5 m. Calculate the work done by this force during this displacement. [Ans (as printed in the book): 8 N]

[!TLDR] Work done = area of the trapezium under the linearly-decreasing force-displacement graph = average force x displacement = 4 N x 2 m = 8 J (the book prints '8 N', a unit typo -- the physically correct unit fo …

The force decreases LINEARLY from 5 N (at s=3 m) to 3 N (at s=5 m) over a displacement of $5-3=2$ m. Since the force-displacement graph is a straight (linearly decreasing) line, the work done equals the AREA of the resulting trapezium under the graph (4.5.5), which equals the AVERAGE force multiplied by the displacement: $W=\left(\frac{5+3}{2}\right)\times(5-3)=4\times2=8$ J. This can be checked by integration too: if $F=5-1\cdot(s-3)=8-s$ (linear, matching F=5 at s=3 and F=3 at s=5), then $W=\int_3^5(8-s)\,ds=\left[8s-\frac{s^2}{2}\right]_3^5=(40-12.5)-(24-4.5)=27.5-19.5=8$ J, confirming the trapezium-area shortcut. The BOOK's own printed answer is '8 N', but work done must have the UNIT OF ENERGY (joules), not force (newtons) -- this is a …

Recognise the linearly-varying-force case allows the trapezium-area shortcut (average force x displ …

Copying the book's printed unit 'N' for what is actually a quantity of WORK (energy), which must …