Numerical Problems Based on Newtons Third Law and Motion in a Lift for Class 11 Physics

Mass of the elevator, $m = 5000\text{ kg}$; Tension, $T = 50,000\text{ N}$. Let’s take $g = 9.8\text{ ms}^{-2}$.

1. Upward acceleration ($a$):
For upward motion, the equation is $T – mg = ma$.

2. Distance traveled ($s$):
Initial velocity, $u = 0$; Time, $t = 10\text{ s}$. Using the second equation of motion:

Mass of the woman, $m = 50\text{ kg}$ (since weight is $50\text{ kgf}$). The reading on the weighing machine is equal to the normal reaction force ($R$).

(i) Moving upwards with uniform velocity ($v = 5\text{ ms}^{-1}$):
Since velocity is uniform, acceleration $a = 0$.

(ii) Moving downwards with uniform acceleration ($a = 1\text{ ms}^{-2}$):
The apparent weight decreases when accelerating downwards. The equation is $mg – R = ma \implies R = m(g – a)$.

Mass of the man, $m = 75\text{ kg}$; Upward acceleration, $a = 2.0\text{ ms}^{-2}$.

The force exerted by the floor is the apparent weight or normal reaction ($R$). For upward acceleration, $R – mg = ma \implies R = m(g + a)$.

To convert this force into $\text{kgf}$, we divide by the value of $g$:

Mass, $m = 49\text{ kg}$. To find the answer in $\text{kgf}$, we calculate the normal reaction ($R$) in Newtons and divide by $g$ (where $g = 9.8\text{ ms}^{-2}$).

(i) Rising with $a = 1.2\text{ ms}^{-2}$:

(ii) Going down with $a = 1.2\text{ ms}^{-2}$:

(iii) Falling freely ($a = g$):

(iv) Uniform velocity ($a = 0$):

Mass, $m = 15\text{ kg}$. The reading of the spring balance gives the apparent weight (Tension, $T$). To convert Newtons to $\text{kgf}$, divide by $g = 10\text{ ms}^{-2}$.

(i) Ascending with $a = 2\text{ ms}^{-2}$:

(ii) Descending with $a = 2\text{ ms}^{-2}$:

(iii) Descending with constant velocity ($a = 0$):

Mass, $m = 10\text{ kg}$; Upward acceleration, $a = 2\text{ ms}^{-2}$.

When the hand is moved upwards, it acts exactly like an elevator accelerating upwards. The equation for the tension ($T$) in the string is:

Maximum Tension, $T_{max} = 72\text{ kg wt} = 72g\text{ N}$.

Mass of the person, $m = 96\text{ kg}$. Let $a$ be the downward acceleration. The equation of motion for downward descent is:

Taking $g = 9.8\text{ ms}^{-2}$:

This is the minimum downward acceleration required to ensure the tension does not exceed the maximum limit of $72\text{ kg wt}$.