# 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?

Jan 8, 2016

${\text{630 kg/m}}^{3}$

#### 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 ${\text{1000 kg/m}}^{3}$ and the 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

${V}_{\text{water" = V_"person" + V_"jacket}}$

Plug in your values to get

${V}_{\text{water" = 6.7 * 10^(-2)"m"^3 + 2.7 * 10^(-2)"m}}^{3}$

${V}_{\text{water" = 9.4 * 10^(-2)"m}}^{3}$

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 $\text{77 kg}$, you can say that the mass of the jacket will be equal to

${m}_{\text{displaced water = "m_"jacket" + m_"person}}$

${m}_{\text{jacket" = "94 kg" - "77 kg" = "17 kg}}$

Since density is defined as mass per unit of volume, use the known volume of the jacket to determine the mass of ${\text{1 m}}^{3}$

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

$\rho = \textcolor{g r e e n}{{\text{630 kg/m}}^{3}}$