A person with a mass of 59 kg climbs up to a diving board 4 meters above a swimm

Question

A person with a mass of 59 kg climbs up to a diving board 4 meters above a swimming pool (20mX20mX10m) carrying a 1 kg gold bar. He then walks 1m towards the edge of the diving board and drops the gold bar into the water below.

a.) How much work does the person do getting to the diving board? How much work while walking to the edge of the board?

b.) How much potential energy does the diver have on the diving board with respect to the surface of the water (assume height of 4m). How much potential energy does the gold brick have if held at chest level of this person (1.5m) over the water?

c.) What is the gold bar's velocity upon entering the water? What law did you use to determine this?

d.) What is the amount of work that the gold bar does on the water when it enters the pool?

e.) If the gold's mass density is 19300 kg/m^3, what is the buoyant force when submerged in water?

f.) At what distance will the bar have 0 net force (i.e. will no longer sink further down)?

g.) What is the absolute pressure at the bottom of the pool if the pool is 20 m (height) X 20m (width) X10m (depth)?

h.) Which of the above quantities will change if the person throws the bar with an initial velocity of 5 m/s and at an angle of 45 degrees? Why?

Explanation / Answer

a)

W = M*g*h = 59*9.8*4 = 2312.8 J

b)

Ud = M*g*h = 2312.8 J

Ugold = m*g*(h+1.5) = 3180.1 J

c)

v = sqrt(2*g*(h+1.5)

v = 10.4 m/s

d)

W = change in KE = 0.5*1*10.4^2 = 54.1 J

e)

Fb = dwater*V*g = 1000*(1/19300)*9.8 = 0.51 N

f)

at the bottom

g)

P = po + dwater*g*H

P = 10^5 + (1000*9.8*20)

P = 2.96*10^5 Pa

h)

c , d

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