posted by Ryan on .
Holding a glass of water filled to a depth of 7.8 cm you step into an elevator. The elevator moves upward with constant acceleration, increasing its speed from zero to 2.2 m/s in 3.1 s.
Find the change in pressure exerted on the bottom of the glass during the time the elevator accelerates.
I understand that you use the P_2_=P_at_+ p_water_(g+a)h!
I find P_2_ and I attained a difference of pressures to be 820. I keep rechecking my answers and submiting new ones, I just cant seem to understand it!
The pressure at the bottom of the glass changes from
rho(water)*g*h to rho(water)*(g+a)*g
when the elevator is accelerating
I have neglected the atmospheric pressure term, which does not change, and is the same on both sides of the bottom of the glass, anyway.
The pressure increase is rho(water)*a*h
h = 0.078 m
a = (2.2 m/s)/3.1s = 0.71 m/s^2
rho(water) = density = 1000 kg/m^2
I meant to write
<<The pressure at the bottom of the glass changes from
rho(water)*g*h to rho(water)*(g+a)*h>>
but the last letter came out g.
The final answer remains correct. The pressure will be in Pascals, ths standard S.I. (Systeme International) unit.