Friction: A factory robot drops a 10 kg computer onto a conveyor belt running at 3.1 m/s. The materials are such that μ_s = 0.50 and μ_k = 0.30 between the belt and the computer. How far is the computer dragged before it is riding smoothly on the belt? A) 1.6 m B) 0.98 m C) 2.3 m D) 3.0 m

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Quizplus · Accepted Answer

The force of friction is given by:$$f = \mu * n$$where:* f is the force of friction* μ is the coefficient of friction* n is the normal forceIn this case, the normal force is equal to the weight of the computer:$$n = mg = 10 kg * 9.8 m/s^2 = 98 N$$The force of friction is then:$$f = 0.50 * 98 N = 49 N$$The acceleration of the computer is:$$a = f/m = 49 N / 10 kg = 4.9 m/s^2$$The distance the computer is dragged before it is riding smoothly on the belt is:$$d = vt + 1/2 * at^2 = 3.1 m/s * t + 1/2 * 4.9 m/s^2 * t^2$$We can solve for t by setting d = 0:$$0 = 3.1 m/s * t + 1/2 * 4.9 m/s^2 * t^2$$$$0 = 3.1t + 2.45t^2$$$$0 = t(3.1 + 2.45t)$$$$t = 0, -1.27 s$$The negative solution is not physical, so we have:$$t = 0 s$$Therefore, the computer is dragged 0 m before it is riding smoothly on the belt.

Friction: a Factory Robot Drops a 10 Kg Computer onto a Conveyor