Solution:
According to the work-energy theorem, \(W = \Delta K = \frac{1}{2}m(v_f^2 - v_i^2)\). Substituting \(v_i = 10\text{ m/s}\) and \(v_f = 30\sqrt{3}\text{ m/s}\), we obtain \(W = \frac{1}{2} \times 1 \times (2700 - 100) = 1300\text{ J}\).
A body of mass 1 kg travels in a straight line with velocity \(v = \alpha x^{3/2}\) where \(\alpha = 10\text{ m}^{-1/2}\text{ s}^{-1}\). The work done by the net force during its displacement from \(x = 1\text{ m}\) to \(x = 3\text{ m}\) is
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