Solution:
Work done \(W = \int_{0}^{10} (-0.6x^2 + 2.5)dx = [-0.2x^3 + 2.5x]_0^{10} = -175\text{ J}\). From the work-energy theorem, \(W = \frac{1}{2}m(v_f^2 - v_i^2) \implies -175 = \frac{1}{2}(2)(v_f^2 - 400)\text{, giving } v_f = 15\text{ m/s}\).
A force \(F = (-0.6x^2 + 2.5)\text{N}\) displaces a particle of mass 2 kg from \(x = 0\) to \(x = 10\text{ m}\). If velocity of particle at \(x = 0\) was 20 m/s, then velocity at \(x = 10\text{ m}\) will be
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