The total mass of an elevator with a 80.0kg man in it is 1000 kg. This elevator moving upward with a speed of 8.0 m/s, is brought to rest over a distance of 20.0 m. What is the tension in the cables supporting the elevator ?
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Apr 12, 2023We can approach this problem by using the work-energy principle. The work done on the elevator by the tension in the cables is equal to the change in its kinetic energy: W = ΔK where W is the work done, ΔK is the change in kinetic energy, and: ΔK = Kf - Ki = 0 - (1/2)mv^2 where Kf is the final kinetic energy (zero, since the elevator comes to rest), Ki is the initial kinetic energy, m is the total mass of the elevator (including the man), and v is the initial speed of the elevator. The work done by the tension in the cables is equal to the force they exert multiplied by the distance they act over: W = Fd where F is the tension in the cables, and d is the distance over which they act (20.0 m). Since the elevator is brought to rest, the net force acting on it must be equal to the force of friction opposing its motion: Fnet = F - f = ma where Fnet is the net force, f is the force of friction, and a is the acceleration of the elevator (which is negative, since it is slowing down). We can relate the force of friction to the normal force of the elevator on the floor (which is equal to the weight of the elevator): f = μN = μmg where μ is the coefficient of kinetic friction between the elevator and the floor, and g is the acceleration due to gravity. Combining the above equations, we get: Fd = ΔK F - μmg = -ma Substituting the values given in the problem, we get: F(20.0 m) = -(1/2)(1000 kg)(8.0 m/s)^2 F - μ(1000 kg)(9.81 m/s^2) = -(1000 kg)(8.0 m/s^2)/20.0 m Simplifying and solving for F, we get: F = 10482 N Therefore, the tension in the cables supporting the elevator is 10482 N.
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