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? $(g=9.8{\text{m/s}}^2)$

Given that,

Mass m=50kg

Initial speed u=3m/s

Distance covered s=9m

The lift brought to rest by constant acceleration in distance of 9m.

So, final speed v=0

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

Negative sign shows that acceleration is downward

Now, the net acceleration is

a=−(a+g)

a=−(0.5+9.8)

a=−10.3 $m/s^2$

Now, the force is

F=ma

F=50×10.3

F=515N

Hence, the scale reading is 515 N.