An elevator\'s cable snaps and its emergency brakes start gripping at the second

Question

An elevator's cable snaps and its emergency brakes start gripping at the secondary line in order to slow its descent. The elevator and passengers weigh 336 kgs together and fall under the influence of gravity (9.81 m/s2). The coefficient of friction between the brakes and secondary line is 0.85, while the emergency brakes are designed such that the normal force on the line is equal to the weight of the elevator and passengers. How fast is the elevator going after falling 3 stories? (One story is precisely 4.50 meters)

m/s

Explanation / Answer

Mass m = 322 kg Normal force Fn = mg Friction Ff = coeff. of friction * Fn = 0.95 * mg Forces in vertical direction are 1. Weight mg downward 2. Friction Ff = 0.95 * mg upward Take downward direction as positive. Net force F = mg - 0.95 * mg Or F = 0.05 * mg Acceleration a = F/m Or a = 0.05 * mg/m Or a = 0.05 g = 0.05 * 9.81 m/s^2 = 0.491 m/s^2 Initial velocity u = 0 Displacement S = 7 storeis = 7 * 4.50 m = 31.5 m Final velocity v = ? v^2 = u^2 + 2aS Or v^2 = 0 + 2 * 0.491 * 31.5 Or v^2 = 30.9 Or v = sqrt(30.9) = 5.56 m/s Ans: 5.56 m/s

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