Question

Consider a wod raft with length = 5.0m, width 3.0 m and height = 1.0m, with density =600kg/m^3. There is an object of a mass of 500 kg standing on the raft. How deep will it sink into the water of density 1000 kg/m^3?

Answer 1

Answer:

0.63333m

Explanation:

If the wood raft partially sinks, it means the water exerts a force upwards, equal to the weight of the displaced water by the part sank that is in magnitude equal to the weight of the raft and the objects on it but the opposite direction.  There fore, we need to calculate the mass of the raft and add it to the object mass for the total.

$m_r=V_r\rho_r$

Where $V_r=5\times3\times1=15m^3$ is the volume of the raft and$\rho=600kg/m^3$ is the density.

$m_r=(600)(15)=9000Kg$

The total mass is 9500Kg (The raft plus the object), and this is equal to the mass of the water displaced. Now find the volume of water that has this mass:

$V=m/\rho$

$V_w=m_w/\rho_w=\frac{9500}{1000} =9.5m^3$

With this volume you can calculate at what height of the raft the water will be. The base of the raft will be the same, just calculate the new height.

$V=h\times B\\h=V/B=\frac{9.5}{5\times3}=\frac{9.5}{15}=0.63333m$