# An 77-kg person puts on a life jacket, jumps into water, and floats. The jacket's volume is #2.7*10^-2m^3# and is completely under water. The volume of the person's body that is under water is #6.7*10^-2m^3#. What is the density of the life jacket?

##### 1 Answer

#### Explanation:

Since the problem provides you with the *volume* of the jacket, you will have to focus on find its **mass** by using the **total mass** of water displaced by the *person and the life jacket*.

To get the total mass of water, you need to use water's approximate density of **total volume of water** displaced by the person and the life jacket.

Now, the trick here is to realize that the volume of water displaced **by the jacket** will be **equal to** the volume of the jacket. Likewise, the volume of water displaced **by the person** will be equal to the volume of the person.

This means that you can write

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

Plug in your values to get

#V_"water" = 6.7 * 10^(-2)"m"^3 + 2.7 * 10^(-2)"m"^3#

#V_"water" = 9.4 * 10^(-2)"m"^3#

Therefore, the total mass of the water displaced by the person and his life jacket will be

#9.4 * 10^(-2)color(red)(cancel(color(black)("m"^3))) * "1000 kg"/(1color(red)(cancel(color(black)("m"^3)))) = "94 kg"#

Since you know that the person has a mass of *mass of the jacket* will be equal to

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

#m_"jacket" = "94 kg" - "77 kg" = "17 kg"#

Since density is defined as mass per unit of volume, use the known volume of the jacket to determine the mass of

#1 color(red)(cancel(color(black)("m"^3))) * "17 kg"/(2.7 * 10^(-2)color(red)(cancel(color(black)("m"^3)))) = "629.63 kg"#

Therefore, the density of the jacket, rounded to two sig figs, will be equal to

#rho = color(green)("630 kg/m"^3)#