   # . A rectangular block of ice with dimensions 2 m by 3 m by 0.2 m floats on water. A person weighing 600 N wants to stand on the ice. Would the ice sink below the surface of the water? ### Inquiry into Physics

8th Edition
Ostdiek
Publisher: Cengage
ISBN: 9781337515863

#### Solutions

Chapter
Section ### Inquiry into Physics

8th Edition
Ostdiek
Publisher: Cengage
ISBN: 9781337515863
Chapter 4, Problem 30P
Textbook Problem
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## . A rectangular block of ice with dimensions 2 m by 3 m by 0.2 m floats on water. A person weighing 600 N wants to stand on the ice. Would the ice sink below the surface of the water?

To determine

A rectangular block of ice with dimensions 2m by 3m by 0.2m floats on water. A person weighing 600N wants to stand on the ice. Would the ice sink below the surface of the water?

### Explanation of Solution

Given:

Dimensions of rectangular block of ice = 2m by 3m by 0.2m.

Weight of the person standing on the block = 600N.

Concept Used:

Archimedes’ Principle is being used here to check if the ice would sink. The ice would sink if the total downward force overcomes the buoyant force exerted by the water upwards. We need to find both the forces and see if the difference between the buoyant force and the downward force is greater than 600 N (weight of the person standing on the ice).

Downward force is due to the weight of the block of ice which is

Fdown=Dice×V×g

Where we have Dice as the density of ice in kg/m3 which is approximately 900kg/m3 V is the volume of the rectangular block of ice.

Acceleration due to gravity, g = 9.8 m/s2.

Now the buoyant force

Fw = Weight density of the water displaced by the ice × Volume

Here volume of water displaced is same as the volume occupied by the ice block. So, expression is

Fw=Dw×V

Dwis the weight density of water = 1000 kg/m3 × g = 980 kg/m2 s2

We have to find if Fw−Fd°wn >600 N

If it’s true, then the person standing on ice won’t sink. Else he will sink.

Calculation:

Volume of the ice block

V=2×3×0

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