The density of a 6 foot tall man that weighs 840. Newtons fluctuates from a valu
ID: 1485136 • Letter: T
Question
The density of a 6 foot tall man that weighs 840. Newtons fluctuates from a value slightly less than the density of water when the man completely inhales to a value slightly more than the density of water when the man completely exhales.
Note: Assume that the water pressure does not affect the volume of the man.
1.) The man completely inhales and jumps into the deep end of a pool with a depth of 9 feet. The man swims to the bottom of a pool and stands erect on the bottom of the pool for a moment at rest. The man then accelerates upward due to an unbalanced vertical force (the buoyant force of the water acting on the man is greater than the weight of the man); the man moves upward 3 feet (0.914m) in 5.59 seconds when his head breaks the surface of the water.
A.) Calculate the net acceleration of the man with his breath completely inhaled while totally submerged in the water.
Hint: Use a kinematics equation.
B.) Calculate the buoyant force of the water acting on the man with his breath completely inhaled.
Note: This is an alternative method to calculate the buoyant force of the fluid acting on an object that uses Newton's Second Law.
Note: Enumerate all force terms on the summation side of Newton's Second Law.
Explanation / Answer
Data Given -
displacement, s = 0.914 m
time taken, t = 5.59 sec
as man was initially at rest , u = 0
s = ut + 0.5 at2
0.914 = 0.5*a*5.592
a = 0.0585 m/sec2
using newton's law of motion -
F = Ma
B - W = Ma
B = W + Ma
B = 840 + [840/9.8]*0.0585
B = 845.01 N