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Chapter 15: Problem 52

A 55 -kg swimmer climbs onto a Styrofoam block of density \(160 \mathrm{kg} / \mathrm{m}^{3} .\) If the water level comes right to the top of the Styrofoam, what's the block's volume?

Short Answer

Expert verified
The volume of the Styrofoam block is \(3.44 m^3\).

Step by step solution

01

Calculate the Weight of the Swimmer

First, you need to calculate the weight of the swimmer. Weight can be calculated by multiplying mass by the acceleration due to gravity. The swimmer's mass is given as 55 kg and the acceleration due to gravity is approximately \(9.81 m/s^2\). So the weight \(W\) of the swimmer will be \(55 kg \times 9.81 m/s^2 = 539.55 N\).
02

Calculate the Buoyant Force

In this case, the buoyant force is equal to the weight of the swimmer because the water level comes exactly to the top of the Styrofoam block. So the buoyant force \(F_b\) is also \(539.55 N\).
03

Calculate the Volume of the Block

According to Archimedes' principle, the buoyant force is equal to the weight of the fluid displaced by the object. The weight of the fluid is the product of the volume and the density. Thus, the volume \( V \) of the Styrofoam block can be calculated by dividing the buoyant force by the density of the Styrofoam and the acceleration due to gravity. So, \( V = 539.55 N / (160 kg/m^3 \times 9.81 m/s^2) = 3.44 m^3\).

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