Question

A 50 kg boy stands on a platform spring scale in a lift that is going down with a constant speed $3\text{m/s}$. If the lift is brought to rest by a constant deceleration in a distance of $9\text{m}$, what does the scale read (in Newtons) during this period ? $(g=9.8{\text{m/s}}^2)$

Answer 1

Using the 3rd equation of motion
$v^2=u^2+2as$
Taking downward direction as positive,
$0=(3{)}^2-2(a\left)\right(9)$
$\therefore a=\frac{1}{2}=0.5{\text{m/s}}^2$ (upwards)
$N=m(g+a)$
$=50(9.8+0.5)=515\text{N}$