The mass of a lift is 5600 kg. The lift starts from rest and descends with uniform acceleration for 8 s until it reaches a speed of V ms–1. The tension in the lift cable is 50 400 N.
Show that the magnitude of the acceleration of the lift is 0·8 ms–2 and Find the value of V.
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The mass of the lift is 5600 kg. It starts from rest and descends with uniform acceleration for 8 s until it reaches a speed of V m/s. The tension in the lift cable at this instant is 50400 N.
The tension in the lift is equal to the net force exerted by the Earth due to the gravitational force of attraction. This is equal to 50400 N, if the lift were restrained by the cable and prevented from moving downwards the tension in the cable would be 5600*9.8 = 54880 N. If the acceleration of the lift is given by A m/s^2. A*5600 = 54880 - 50400 = 4480. A = 4480/5600 = 0.8 m/s^2.
The initial velocity of the lift was 0 m/s and it accelerated downwards for 8 s at 0.8 m/s^2. The final velocity V of the lift is equal to 0.8*8 = 6.4 m/s
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