Hi. Can someone please check these two answers? (a) seems very low and (b) seems very high to me.
A 0.22 micro-F capacitor is charged by a 1.5-V battery. After being charged, the capacitor is connected to a small electric motor. Assuming 100% efficiency, (a) to what height can the motor lift a 5.0-g mass? (b) What initial voltage must the capacitor have if it is to lift a 5.0-g mass through a height of 1.0 cm?
C = 0.22 micro-F = 0.22 x 10^-6 F
V = 1.5 V
m = 5.0 g = 5.0 x 10^-3 kg
h = ?
Q = CV = 3.3 x 10^-7
U = (1/2)QV = 2.5 x 10^-7 J
ANSWER: U = mgy , y = U/mg = 5.1 x 10^-6 m
V = ?
m = 5.0 x 10^-3 kg
C = 0.22 x 10^-6
h = 1 cm = 0.01 m
U = mgy = 4.9 x 10^-4 J
ANSWER: U = (1/2)QV , V = U /(1/2)Q = 2970 V
Your work for (a) is certainly correct.
Your work for (b) is certainly correct.
Whoever wrote the question certainly didn't use realistic examples. Although (a) is a bit realistic if you think of MEMS (Micro-electro-mechanical systems). Moving a micro-mechanical beam 5.1um ...
The solution includes a 150-200 word explanation.