A 3 kg object moving at 8 m/s comes to rest after covering 12 m on a rough surface. What is the magnitude of the average frictional force acting on it ?
8 N
Using the work-energy theorem, the work done by friction equals the loss in kinetic energy of the object.
Initial kinetic energy: \(KE = \dfrac{1}{2}mv^2 = \dfrac{1}{2} \times 3 \times 8^2 = \dfrac{1}{2} \times 3 \times 64 = 96\, J\). Final kinetic energy is zero since the object comes to rest.
So the frictional force does work: \(F \times d = 96\), giving \(F \times 12 = 96\), so \(F = \dfrac{96}{12} = 8\, N\).
Therefore, the magnitude of the average frictional force is 8 N.
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