Question

total mass of an elevater with 80 kg man in it is 1000 kg it is brought to rest over a distance of 16m then during retardation the tension in the supporting cable force exerted by man on the floor is respectively​

Answer 1

Solution :

• The elevator having an initial upward speed of 8 m/sec is brought to rest within a distance of 16 m. Hence

v² = u² + 2as

➨ 0² = 8² + 2 * a * 16

➨ 0 = 64 + 32a

➨ 32a = - 64

➨ a = - 64/32

➨ a = - 2 m/s²

Resultant upward force on elevator = T – mg.

According to Newton's law

T – mg = ma

➨ T = mg + ma

➨ T =m(g + a)

➨ T = 1000 (9.8 – 2)

➨ T = 7800 N

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• Let P be the upward force exerted on the man by the elevator floor. If m´ be the mass of the man, then

Weight of the man acting downward = m´g

Upward force on the man = P – m´g.

According to Newton's law

P – m´g = m´a

➨ P = m´(a + g)

➨ P = 80 (– 2 + 9.8)

➨ P = 624 N