Consider a system of a cliff diver and the earth. The gravitational potential en
ID: 1536491 • Letter: C
Question
Consider a system of a cliff diver and the earth. The gravitational potential energy of the system decreases by 22,000 J as the diver drops to the water from a height of 35.0 m. Determine her weight in newtons. Consider a system of an 85.0-kg man, his 15.5-kg dog, and the earth. The gravitational potential energy of the system increases by 2.25 times 10^3 J when the man climbs a spiral staircase from the first to the second floor of an apartment building. If his dog climbs a normal staircase from the same first floor to the second floor, by how much does the potential energy of the system increase? While leaning out a window that is 6.2 m above the ground, you drop a 0.60-kg basketball to a friend at ground level. Your friend catches the ball at a height of 1.9 m above the ground. Determine the following. (a) the amount of work done by the force of gravity on the ball J (b) the gravitational potential energy of the ball-earth system, relative to the ground when it is released J (c) the gravitational potential energy of the ball-earth system, relative to the ground when it is caught J (d) the ratio of the change (PE_f - PE_0) in the gravitational potential energy of the ball-earth system to the work done on the ball by the force of gravityExplanation / Answer
5. As the earth is a really large mass when compared to the man, assume the mass of earth be M
Mass of diver = m
Initial distance between the diver and earth = d
FInal distance between the diver and the earth = d - 35m
Initial PE of the system = -GMm/d
Final PE of the system = -GMm/(d -35)
Change in PE of system = Initial PE - FInal PE = GMm(1/(d-35) - 1/d) = GMm*35/d*(d - 35) = 22,000 J
Now, G is gravitational constant
so Acceleration due to gravity, g = GM/d^2 [ as d is the radius of earth ]
so,
22000 = g*d^2*35m/d(d-35)
22,000 = g*35m/(1 - 35/d)
as 35/d < < < 1
so, 22,000 = 35mg
mg = 628.571 N