The answer to each of the following questions can be written as a numeric coeffi

Question

The answer to each of the following questions can be written as a numeric coefficient times some combination of the variables m, g, and d (representing mass, acceleration due to gravity, and the height of one step respectively). The appropriate combination of variables is indicated. Enter only the numeric coefficient. (Example: If the answer is 1.23mgd, just enter 1.23) Three different objects, all with different masses, are initially resting at the bottom of a set of steps, each with a uniform height d. In this position, the total gravitational potential energy of the three object system is said to be zero. If the objects are then relocated as shown, what is the new total potential energy of the system? 4.60m Number Ug.systcem mgd g,syste 2.21m This potential energy was calculated relative to the bottom of the stairs. If you were to redefine the reference height such that the total potential energy of the system becomes zero, how high above the bottom of the stairs would the new reference height be? In Number original reference height Now, find a new reference height (measured again from the bottom of the stairs) such that the highest two objects have the exact same potential energy

Explanation / Answer

a)
Potential energy will be negative
U = -m1*g*h1 - m2*g*h2 - m3*g*h3
= - m*g*d - 2.21m*g*2d - 4.6m*g*3d
= (-1-4.42-13.8)*(m*g*d)
= -19.22 m*g*d
Answer: -19.22 m*g*d

b)
Let the reference height be at height h
then total potential energy of system= 0
m1*g*(h-d) + m2*g*(h-2d)+m3*g*(h-3d) = 0
m*(h-d) + 2.21m*(h-2d)+4.6m*(h-3d) = 0
mh-md + 2.21mh- 4.42md + 4.6mh - 13.8md = 0
7.81*h = 19.22 d
h=2.46 d
Answer:2.46 d

c)
Let the reference height be at height h
potential energy of 2nd mass = potantial energy of highest mass
m2*g*(h-2d) = m3*g*(h-3d)
2.21m*(h-2d) = 4.6m*(h-3d) =
2.21mh- 4.42md = 4.6mh - 13.8md
9.38d = 2.39 h
h=3.92 d
Answer:3.92 d

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