The only force acting on a \(2\text{ kg}\) body as it moves along the positive x axis has component \(F_x = -6x\text{ N}\), where x is in metre. The velocity of the body at \(x = 3\text{ m}\) is \(8\text{ m/s}\). The velocity of the body at \(x = 4\text{ m}\) is:
Solution:
Applying the work-energy theorem, \(W = \Delta K ⇒ \int_{3}^{4} -6x , dx = \frac{1}{2} m(v_f^2 - v_i^2)\). Integrating gives \(-3(16 - 9) = \frac{1}{2} (2) (v_f^2 - 64)⇒ -21 = v_f^2 - 64\), which yields \(v_f = \sqrt{43} \approx 6.6\text{ m/s}\).
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