A particle of mass m moves along x-direction under the force \(F(x) = F_0\left(1-\dfrac{x}{L}\right)\) for \(0 \leq x \leq L\). If the particle starts from rest at x = 0, then what is the speed (v) of the particle at x = L?
\(v = \sqrt{\dfrac{F_0L}{m}}\)
By the work-energy theorem, the work done by the force as the particle moves from x = 0 to x = L equals its gain in kinetic energy. \(W = \displaystyle\int_0^L F_0\left(1-\dfrac{x}{L}\right)dx = F_0\left[x-\dfrac{x^2}{2L}\right]_0^L = F_0\left(L-\dfrac{L}{2}\right) = \dfrac{F_0L}{2}\). Since the particle starts from rest, \(\dfrac{1}{2}mv^2 = \dfrac{F_0L}{2}\), which gives \(v = \sqrt{\dfrac{F_0L}{m}}\).
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