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Question

Physics Question on laws of motion

The mass of a lift is 2000 kg.2000\, kg. When the tension in the supporting cable is 28000 N,28000\, N, then its acceleration is

A

4 ms−24\, ms^{-2} upwards

B

4 ms−24\, ms^{-2} downwards

C

14 ms−214\, ms^{-2} upwards

D

30 ms−230\, ms^{-2} downwards

Answer

4 ms−24\, ms^{-2} upwards

Explanation

Solution

ΣF=ma\Sigma F = ma
⇒T−mg=ma\Rightarrow T - mg = ma
a=T−mgm=4 m/s2a =\frac{ T - mg }{ m }=4\, m / s ^{2}