To solve this problem we will use the concepts related to Torque as a function of the Force in proportion to the radius to which it is applied. In turn, we will use the concepts of energy expressed as Work, and which is described as the Torque's rate of change in proportion to angular displacement:

Where,
F = Force
r = Radius
Replacing we have that,



The moment of inertia is given by 2.5kg of the weight in hand by the distance squared to the joint of the body of 24 cm, therefore


Finally, angular acceleration is a result of the expression of torque by inertia, therefore



PART B)
The work done is equivalent to the torque applied by the distance traveled by 60 °° in radians
, therefore



Answer:
Sorry don't know the answer
1) The total mechanical energy of the rock is:

where U is the gravitational potential energy and K the kinetic energy.
Initially, the kinetic energy is zero (because the rock starts from rest, so its speed is zero), and the total mechanical energy of the rock is just gravitational potential energy. This is equal to

where

is the mass,

is the gravitational acceleration and

is the height.
Putting the numbers in, we find the potential energy

2) Just before hitting the ground, the potential energy U is zero (because now h=0), and all the potential energy of the rock converted into kinetic energy, which is equal to:

where v is the speed of the rock just before hitting the ground. Since the mechanical energy of the rock must be conserved, then the kinetic energy K before hitting the ground must be equal to the initial potential energy U of the rock:

3) For the work-energy theorem, the work W done by the gravitational force on the rock is equal to the variation of kinetic energy of the rock, which is:
<span>Because they occur at an atomic level, changing the actual structure of the thing.
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