A body of mass 11kg falls freely from rest from a height of 15m above the ground. 3m above the ground it breaks through a thin roof before continuing its fall.

(a) given that the work done in breaking through the roof is 130 J. Calculate the velocity of the body after it emerges from the roof.
(b) calculate the velocity of the body as it hits the ground.
(c) An identical body is dropped from the same height through the hole in the roof made by the first body . Ccalculate the velocity of the b0dy sas it hits the ground
(d) which mechanical principle did you apply when answering the above.

a. KE = 0.5*M*V^2 = 130 J.

0.5*11*V^2 = 130
V^2 = 23.64
V = 4.86 m/s.

b. V^2 = Vo^2 + 2g*h = 4.86^2 + 19.6*3 =
82.4
V = 9.1 m/s

c. V^2 = Vo^2 + 2g*h = 0 + 19.6*15 = 294
V = 17.15 m/s.

(a) To calculate the velocity of the body after it emerges from the roof, we can use the principle of conservation of energy. The work done in breaking through the roof is equal to the change in kinetic energy of the body. We can set up the equation as follows:

Work done = Change in kinetic energy
130 J = (1/2) * mass * (velocity^2) - 0 [since the body is initially at rest]

Since mass = 11 kg, we can rearrange the equation to solve for velocity:

130 J = (1/2) * 11 kg * (velocity^2)
260 J = 11 kg * (velocity^2)
Divide both sides by 11 kg:
velocity^2 = 260 J / 11 kg
velocity^2 = 23.64 m^2/s^2
Taking the square root of both sides:
velocity = √(23.64 m^2/s^2)
velocity ≈ 4.863 m/s

Therefore, the velocity of the body after it emerges from the roof is approximately 4.863 m/s.

(b) To calculate the velocity of the body as it hits the ground, we can again use the principle of conservation of energy. The potential energy of the body at the starting height is converted to kinetic energy at the ground level.

Potential energy at height = Kinetic energy at ground level
mgh = (1/2) * mass * (velocity^2)
11 kg * 9.8 m/s^2 * 15 m = (1/2) * 11 kg * (velocity^2)
1617 J = 5.5 kg * (velocity^2)
velocity^2 = 1617 J / 5.5 kg
velocity^2 ≈ 294.182 m^2/s^2
Taking the square root of both sides:
velocity ≈ √(294.182 m^2/s^2)
velocity ≈ 17.135 m/s

Therefore, the velocity of the body as it hits the ground is approximately 17.135 m/s.

(c) Since the second body is identical and it is dropped from the same height, it will have the same potential energy at the starting height. Therefore, we can use the same equation as in part (b) to calculate the velocity of the body as it hits the ground:

Potential energy at height = Kinetic energy at ground level
mgh = (1/2) * mass * (velocity^2)
11 kg * 9.8 m/s^2 * 15 m = (1/2) * 11 kg * (velocity^2)
1617 J = 5.5 kg * (velocity^2)
velocity^2 = 1617 J / 5.5 kg
velocity^2 ≈ 294.182 m^2/s^2
Taking the square root of both sides:
velocity ≈ √(294.182 m^2/s^2)
velocity ≈ 17.135 m/s

Therefore, the velocity of the second body as it hits the ground is also approximately 17.135 m/s.

(d) The mechanical principle applied in answering the above calculations is the principle of conservation of energy. This principle states that energy cannot be created or destroyed; it can only be transferred or converted from one form to another. In this case, the potential energy of the body at a certain height is converted into kinetic energy as it falls freely. By equating the initial potential energy to the final kinetic energy, we can solve for the final velocity of the body.