A man of 80kg stands on a spring scale in an elevator during the first 10seconds of motion from rest. the tension T in the cable is 9300N. find the reading R of the scale in newtons during the interval and upward velocity v on the elevator at the end of the 6seconds. the total mass of the elevator man and scale is 800kg.

Answer 1

Given:
Total mass (M) = 800 kg
Tension in the cable (T) = 9300 kg ms⁻²
Mass of Man (m) = 80 kg

* Then equilibrium of forces in vertical direction will be:
T-Mg=Ma
Where, g= acceleration due to gravity
and a= acceleration due to upward force(acceleration of elevator)

9300-(800×10)= 800a
a= 1.625 ms⁻²

* For Reading of the scale :
R= m× total acceleration
R= m(g+a)
R= 80(10+1.625)= 80×2.625
= 210 ms⁻² = 210Newtons

* For upward velocity after 6 seconds
final velocity= initial velocity +(acceleration×time)
v=u+at
v= 0+1.625×6 = 9.725 ms⁻¹
velocity at the end of six seconds will be 9.725 ms⁻¹

[Note: g=10ms⁻² was taken for simplicity. Different values will give different answers]