A little aluminum boat (mass of 14.50 g) has a volume of 450.00 cm3. The boat is placed in a small pool of water and carefully filled with pennies. If each penny has a mass of 2.50 g.....

Question

A little aluminum boat (mass of 14.50 g) has a volume of 450.00 cm3. The boat is placed in a small pool of water and carefully filled with pennies. If each penny has a mass of 2.50 g, how many pennies can be added to the boat before it sinks?

Answer 1

If the water level is at the top of the boat (just before it sinks), the buoyant force upwards from the water (the weight of water displaced) must be equal to the weight of the boat.

Suppose we place x pennies.

Mass of the boat in g - 14.50 + 2.50 x

Mass of the boat (kg) - (14.50 + 2.50x) / 1000

Weight of the boat (N) = mg = (14.50 + 2.50x) x 9.81 / 1000

Volume of water displaced = 450 cm³ = 0.45 l

I l water has mass 1 kg

So mass of water displaced = 0.45 kg

Weight of water displaced = 0.45 x 9.81

So (14.50 + 2.50 x ) x 9.81 / 1000 = 0.45 x 9.81

14.50 + 2.50 x = 1000 x 0.45 = 450

x = (450 - 14.50) / 2.50 = 174.2 pennies, so answer is 174

Mike