# An 81-kg person puts on a life jacket, jumps into water, and floats. The jacket's volume is #3.1*10^-2m^3# and is completely under water. The volume of the person’s body that is under water is #6.2*10^-2m^3#. What is the density of the life jacket?

##### 1 Answer

Density:

#### Explanation:

The idea here is that you need to find the *mass* of the life jacket by using the *total mass of water* displaced by the person and his life jacket.

The volume of water displaced by the jacket is equal to the volume of the jacket *under water*. Likewise, the volume of water displaced by *the person* is equal to the volume of the person *under water*.

The **total volume of water** displaced by the person and his life jacket will be

#V_"water" = V_"jacket" + V_"person"#

#V_"water" = 3.1 * 10^(-2)"m"""^3 + 6.2 * 10^(-2)"m"""^3 = 9.3 * 10^(-2)"m"""^3#

If you take water's density to be equal to *by the person and his jacket* will be

#rho = m_"water"/V_"water" implies m = rho_"water" * V_"water"#

#m_"water" = 1000"kg"/color(red)(cancel(color(black)("m"^3))) * 9.3 * 10^(-2)color(red)(cancel(color(black)("m"^3))) = "93 kg"#

Since the mass of the person and his jacket will be equal to

#m_"water" = m_"person" + m_"jacket"#

you can find the mass of the jacket

#m_"jacket" = m_"water" - m_"person" = "93 kg" - "81 kg" = "12 kg"#

The density of the jacket will thus be

#rho = m_"jacket"/V_"jacket" = "12 kg"/(3.1 * 10^(-2)"m"""^3) = "387.1 kg/m"""^3#

Rounded to two sig figs, the answer will be

#rho_"jacket" = color(green)("390 kg/m"""^3)#