2. Am 80.0 kg astronaut is training for accelerations that he will experience upon re-entry to earth’s gravity from space. He is
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Answer:
a) V = 0.085 m^3
b) m = 17 kg
Explanation:
1) Data given
mb = 68 kg (mass for the block)
20% of the block volume is floating
100-20= 80% of the block volume is submerged
2) Notation
mb= mass of the block
Vw= volume submerged
mw = mass water displaced
V= total volume for the block
3) Forces involved (part a)
For this case we have two forces the buoyant force (B), defined as the weight of water displaced acting upward and the weight acting downward (W)
Since we have an equilibrium system we can set the forces equal. By definition the buoyant force is given by :
B = (mass water displaced) g = (mw) g (1)
The definition of density is :
If we solve for mw we got (2)
Replacing equation (2) into equation (1) we got:
(3)
On this case Vw represent the volume of water displaced = 0.8 V
If we replace the values into equation (3) we have
0.8 ρ_w V g = mg (4)
And solving for V we have
V = (mg)/(0.8 ρ_w g )
We cancel the g in the numerator and the denominator we got
V = (m)/(0.8 ρ_w)
V = 68kg /(0.8 x 1000 kg/m^3) = 0.085 m^3
4) Forces involved (part b)
For this case we have bricks above the block, and we want the maximum mass for the bricks without causing it to sink below the water surface.
We can begin finding the weight of the water displaced when the block is just about to sink (W1)
W1 = ρ_w V g
W1 = 1000 kg/m^3 x 0.085 m^3 x 9.8 m/s^2 = 833 N
After this we can calculate the weight of water displaced before putting the bricks above (W2)
W2 = 0.8 x 833 N = 666.4 N
So the difference between W1 and W2 would represent the weight that can be added with the bricks (W3)
W3 = W1 -W2 = 833-666.4 N = 166.6 N
And finding the mass fro the definition of weight we have
m3 = (166.6 N)/(9.8 m/s^2) = 17 Kg