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202211241033 Solution to 1970-HL-PHY-I-1

Show that, when an object of mass m is weighed by a spring-balance suspended from the ceiling of a lift moving upwards with acceleration a, the reading W' in the spring-balance is given by m(g+a), where g is the acceleration due to gravity. W' is called the apparent weight of the object, and mg the true weight.

It is known that a passenger in a lift will experience discomfort if his appparent weight exceeds his true weight by more than one sixth, or falls short of his true weight by more than one eighth. A lift in a hotel is designed to transport passengers from the ground floor direct to the top floor in the shortest possible time without causing discomfort. Find this time if the distance between the ground floor and the top floor is 100 metres.

Roughwork.

By Newton’s 2nd law,

\begin{aligned} F_{\textrm{net}}& = ma \\ W'-mg & = ma \\ W' & = m(g+a) \\ \end{aligned}

By boundary conditions,

\begin{aligned} & \quad \enspace \begin{cases} \displaystyle{\frac{W'-W}{W}\leqslant \frac{1}{6} }\\ \displaystyle{\frac{W-W'}{W}\leqslant \frac{1}{8} }\\ \end{cases} \\ & \Rightarrow \frac{8W}{9} \leqslant W' \leqslant \frac{7W}{6} \\ & \Rightarrow \frac{8}{9}mg\leqslant m(g+a)\leqslant \frac{7}{6}mg \\ & \Rightarrow -\frac{g}{9} \leqslant a\leqslant \frac{g}{6} \\ & \Rightarrow |a|\leqslant \frac{g}{9} \\ \end{aligned}

Take upward positive and let g=10\,\mathrm{m\,s^{-2}},

\begin{aligned} s & = ut+\frac{1}{2}at^2 \\ (100) & = (0)t + \frac{1}{2}\bigg(\frac{10}{9}\bigg) t^2 \\ t & = 6\sqrt{5}\,\mathrm{s} \\ \end{aligned}

This problem is not to be attempted.