Question

Your doctor has told you to watch your weight. Of course, your weight comprises two factors, so you assume that she wants you to watch both your mass and the local value g. Which explains why you are carrying a small mass m on an elastic band wherever you go. Most of the time, the elastic band is stretched by a distance of 2.1 cm. However, when you have finished lunch, you go into a very tall building and take the high speed lift (or elevator depending on where you live). Soon after the doors close, you notice that the band (which has linear elastic properties) is stretched by a total distance of 3.5 cm.

a) At what rate is the lift accelerating?

b) Your doctor will want to know the factor by which your weight has increased after that lunch. Suppose that you stand on scales in the lift while it is accelerating as described above. According to the scales, your weight has increased by a factor of _____ .

Answer 1

$\Delta x = 2.1cm = 2.1*10^{-2}m$

$\text {for the body with an elastic band under normal conditions}$

$\text{according to Hooke's law}$

$F = k\Delta x$

$P = F = mg$

$\Delta x= \frac{mg}{k}=2.1*10^{-2}m$

$a)$

$\text{for an body in an elevator:}$

$\Delta x_1 = 3.5cm = 3.5*10^{-2}m$

$F_1 = k\Delta x$

$P_1 = F_1 = m(g-a)$

$\Delta x_1= \frac{m(g-a)}{k}=3.5*10^{-2}m$

$\frac{\Delta x_1}{\Delta x}=\frac{3.5*10^{-2}}{2.1*10^{-2}}\approx1.67$

$\frac{\Delta x_1}{\Delta x}= \frac{\frac{m(g-a)}{k}}{\frac{mg}{k}}=\frac{g-a}{g}=1-\frac{a}{g}$

$a = -0.67*g = -0.67*9.8= -6.57 \frac{m}{s^2}$

$\text{Acceleration minus sign means acceleration is directed }$

$\text{upwards opposite to gravity}$

$\text{Answer: }a = -6.57 \frac{m}{s^2}$

$b)$

$\text{From expression (1)}$

$1-\frac{a}{g}= 1.67$

$1-\frac{a}{g} = \frac{g-a}{g} = \frac{m_p(g-a)}{m_pg} =\frac{P_1}{P}$

$m_p -\text {mass of pacient}$

$\frac{P_1}{P} = 1.67$

$\text{Answer:}$

$\text{According to the scales, your weight has increased by a factor of } 1.67$