Lift please

Classical Mechanics Level 2

An elevator starts from rest with a constant upward acceleration. It moves 2 m in the first 0.6 s. A passenger in the elevator is holding a 3 kg package by a vertical string. The tension in the string during acceleration is (Take g = 9.8 m/s2)

62.7N 60.7N 61.7N 63.7N

Sanjay Balaji by Sanjay Balaji , 21, India

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4 solutions

Johan Kurniawan
Apr 15, 2014

we know the formula, s=v0 . t+1/2 . a . t^2

2 = 0+1/2 . a . (0.6)^2

a=100/9

F=m . a

T-W=m . a

T-(3 x 9.8)=3 . 100/9

T-(29.4)=100/3

T=33,3+29.4

T=62.7 N

6 Helpful 0 Interesting 0 Brilliant 0 Confused
Sudipan Mallick
Apr 28, 2014

from free body diagram, T-mg=ma T=m(g+a) putting the values:- T=62.7 N

1 Helpful 0 Interesting 0 Brilliant 0 Confused

great!

Arijit Banerjee - 7 years, 1 month ago

Hey yo,

as for this,

as you know as they are moving upwards,

T(package) + W(package) = 0, by newton's 3rd law,

T(package) = W(package),

as for the linear acceleration,

by applying s = ut + 0.5 at^(2)

a = (s - ut ) / 0.5t^2 = ( 2 - 0) / ( 0.5x(0.6)^2) = 2 / 0.18 = 100 / 9 m/s/s

therefore,

T(package) = m(a+g)=3(100/9 + 9.8) = 62.7N

thanks....

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Mayank Holmes
Apr 21, 2014

there is only one point which may bewilder your minds!........ and that's the acceleration which comes out to be 11.11 ....... now that's the acceleration of the lift and not the person. the lift is supported so that it does not fall freely under gravity. so the acceleration of the lift= 11.1111
now, the acceleration of that person is equal to ((11.1111 + 9.8) =20.9111) so the weight of the package = m g = 3 20.91 = 62.73(approx.)

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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