Question

A small metal sphere weighs 0.34 N in air and has a volume of 13 cm3 . What is the acceleration of the sphere as it falls through water

Answer 1

To solve this problem we will apply the concepts related to the balance of forces. Said balance will be given between buoyancy force and weight, both described as derived from Newton's second law, are given as

Buoyancy force

$F_B = V\rho g$

Here,

V = Volume

$\rho$=Density of air

g = Acceleration due to gravity

Weight

$F_W = mg$

m = mass

g = Gravity

Our values are given as,

$\text{Weight of the sphere} = W = 0.34 N$

$\text{Volume} = V = 13 cm^3 = 13*10^{-6}m^3$

$\text{density of air} =\rho =1.29kg/m^3$

$\text{gravity}= g = 9.8 m/s^2$

Then,

$F = V\rho g$

Replacing,

$F = (13*10^{-6}m^3 )(1.29Kg/m^3)( 9.8 m/s^2) = 1.6434* 10^{-4} N$

Now net force is ,

$F_{net} = mg - F$

Mass of the sphere is

$m = \frac{W}{g} = \frac{0.34N}{9.8m/s^2} = 0.03469 kg$

Now acceleration of the sphere is

$a = \frac{F_{net}}{m}$

$a = \frac{( 0.34 N)- (1.6434* 10^{-4} N)}{0.03469 kg}$

$a = 9.822m/s^2$

Therefore the acceleration of the sphere as it falls through water is $9.822m/s^2$