Question

A 4000 kg lift is accelerating upwards. The tension in the
supporting cable is 48000 N. If g = 10ms⁻² then the
acceleration of the lift is
(a) 1 ms⁻² (b) 2 ms⁻²(c) 4 ms⁻² (d) 6 ms⁻²

Answer 1

Question:-

A 4000 kg lift is accelerating upwards. The tension in the

supporting cable is 48000 N. If g = 10ms⁻² then the

acceleration of the lift is

(a) 1 ms⁻² (b) 2 ms⁻²(c) 4 ms⁻² (d) 6 ms⁻²

${\red{\underline{\underline{\bold{Answer:-}}}}}$

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Mass of elevator = 4000 kg

Weight of elevator = Mass × gravity

Weight = 4000 × 9.8

Weight = 39200 N

Upward Force applied = 48000 N

F(net) = 48000 - Weight

F(net) = 48000 - 39200

F(net) = 8800 N

Also, F(net) = mass (m) × acceleration (a)

(a)

A=fnet/m

A=8800/4000

a = 2.2 m/s^2

Now,

Initial velocity (u) = 0 m/s

Time = 3 seconds

H = ?

h=ut+1/2 a ×t²

H= 0+1/2×2.2 ×3²

H = 9.9 m

Hence, the elevator is raised by 9.9 m.

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hope it helps you.