Question

A man of mass $50\text{kg}$ is in a lift at ground floor. The man is now on a weight machine. For first $10\text{s}$, reading of weighing machine is $60\text{kg-wt}$. For next $10\text{s}$, it reads $50\text{kg-wt}$ and for rest of the journey till top floor, reading of weighing machine is $30\text{kg-wt}$. Assume that initial and final velocity of lift is zero. Total distance travelled by the lift is $50x\text{m}$. The value of $x$ is _______.

$(g=10{\text{m/s}}^2)$

Answer 1

Let us work from the lift frame of reference.

For first $10\text{s}$:

Reading of machine is $60\text{kg-wt}=60g\text{N}$.

So,
$N=mg+ma\left[ma\text{is downward pseudo force}\right]$

$\Rightarrow 60g=50g+50a$

$\Rightarrow a=2{\text{m/s}}^2$

$\therefore$ Distance travelled,

$S_1=\frac{1}{2}a{t}^2=\frac{1}{2}\times 2\times {10}^2=100\text{m}...\left(i\right)$

Velocity after $10\text{s}$, $v=at=2\times 10=20\text{m/s}$

For next $10\text{s}$:

Reading of machine is $50\text{kg-wt}=50g\text{N}$. In this case, the reading of machine is equal to the weight of man.
$\Rightarrow a=0$ or velocity of man or lift is constant, $i.e.v=20\text{m/s}$

$\therefore$ Distance travelled during this time,
$S_2=vt=20\times 10=200\text{m}...\left(ii\right)$

For the rest of the journey,
$v=0,u=20\text{m/s}$

Weighing machine's reading is $30\text{kg-wt}=30g\text{N}$

So,

$N+ma=mg\left[ma\text{is upward pseudo force}\right]$


$\Rightarrow 30g+50a=50g$

$\Rightarrow a=4{\text{m/s}}^2$

So, distance travelled during this time,

$v^2=u^2+2a{S}_3$

$\Rightarrow S_3=\frac{u^2}{2a}=\frac{400}{2\times 4}=50\text{m}...\left(iii\right)$

Total distance travelled,
$S=S_1+S_2+S_3$
$\Rightarrow S=100+200+50=350\text{m}\left[\text{From}\right(i),(ii\left)\text{and}\right(iii\left)\right]$

$\Rightarrow S=50\times 7\text{m}=50x\text{m}$

$\Rightarrow x=7$