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) To calculate the velocity of the body after it emerges from the roof, we need to consider the work-energy principle. The work done on the body is equal to the change in its kinetic energy.
Given information:
Mass of the body, m = 11 kg
Height from which the body falls, h = 15 m
Work done in breaking through the roof, W = 130 J
The work done in breaking through the roof is equal to the potential energy loss of the body when it falls through a height of 3 m. The potential energy loss is given by mgh, where g is the acceleration due to gravity.
So, W = mgh = 11 kg * 9.8 m/s^2 * 3 m
The remaining potential energy of the body after emerging from the roof is (mgh - W), i.e., the potential energy at a height of 3 m above the ground. This potential energy will convert entirely into kinetic energy, given that the body falls freely. Thus,
Potential Energy at 3 m = mgh - W = 11 kg * 9.8 m/s^2 * 12 m
The kinetic energy (KE) of the body is given by (1/2)mv^2, where v is the velocity after emerging from the roof.
Equating the potential energy at 3 m to the kinetic energy, we have:
(1/2)mv^2 = mgh - W
Simplifying and solving for v:
v^2 = 2gh - 2W/m
v^2 = 2 * 9.8 m/s^2 * 12 m - 2 * 130 J / 11 kg
v^2 = 235.27 m^2/s^2 - 23.64 m^2/s^2
v^2 = 211.63 m^2/s^2
Taking the square root of both sides:
v ≈ 14.55 m/s
Therefore, the velocity of the body after it emerges from the roof is approximately 14.55 m/s.
(b) To calculate the velocity of the body as it hits the ground, we can use the principle of conservation of mechanical energy. At the highest point (before breaking through the roof), the total mechanical energy is equal to the sum of the potential energy and the work done in breaking through the roof. This remains constant throughout the motion.
So, at the highest point,
Potential Energy + Work done = Total Mechanical Energy
mgh + W = (1/2)mv^2
Using the same values as before,
11 kg * 9.8 m/s^2 * 15 m + 130 J = (1/2) * 11 kg * v^2
Solving for v:
v^2 = (2 * (11 kg * 9.8 m/s^2 * 15 m + 130 J)) / 11 kg
v^2 = 290.91 m^2/s^2
Taking the square root of both sides:
v ≈ 17.04 m/s
Therefore, the velocity of the body as it hits the ground is approximately 17.04 m/s.
(c) An identical body is dropped from the same height through the hole in the roof made by the first body. The height from which it falls and the work done on the body will remain the same as in the previous calculations. Therefore, the velocity of the second body as it hits the ground will also be approximately 17.04 m/s.
(d) The mechanical principles applied in answering the above questions are the work-energy principle and the conservation of mechanical energy. The work-energy principle relates the work done on an object to its change in kinetic energy. The conservation of mechanical energy states that the total mechanical energy of a system remains constant as long as no external forces do work on it.
To answer the questions, we need to apply the principles of work and energy.
(a) To find the velocity of the body after it emerges from the roof, we can use the work-energy principle. The work done in breaking through the roof can be equated to the change in the body's kinetic energy.
The work done (W) is given as 130 J and can be found using the formula:
W = F * d,
where F is the force applied and d is the distance moved in the direction of the force.
Since the body is in free fall, the only force acting on it is gravity. The force due to gravity is given by:
F = m * g,
where m is the mass of the body (11 kg) and g is the acceleration due to gravity (approximately 9.8 m/s^2).
To calculate the distance moved, we can subtract the distance above the roof (15 m) from the distance above the ground (3 m):
d = 15 m - 3 m = 12 m.
Now, we can calculate the velocity after the body emerges from the roof using the work-energy principle:
W = ΔKE,
130 J = (1/2) * m * v^2,
where v is the velocity of the body after it emerges from the roof.
Rearranging the equation and substituting the given values:
v^2 = (2 * W) / m,
v^2 = (2 * 130 J) / 11 kg,
v^2 ≈ 23.63 m^2/s^2.
Therefore, the velocity of the body after it emerges from the roof is approximately 4.86 m/s (taking the square root).
(b) To calculate the velocity of the body as it hits the ground, we can use the principle of conservation of mechanical energy. At the starting height (15 m above the ground), the body has potential energy which converts into kinetic energy as it falls. When it reaches the ground, all potential energy is converted to kinetic energy.
The potential energy when the body is at height h is given by:
PE = m * g * h,
where h is the height above the reference point (in this case, the ground).
The total mechanical energy (E) remains constant throughout the fall and can be expressed as the sum of the potential energy and kinetic energy:
E = PE + KE.
At the top (initial) position, the body has no kinetic energy, so E = PE.
At the ground (final) position, the potential energy is zero, so E = KE.
Therefore, we can equate the potential energy and kinetic energy at the initial and final positions:
m * g * h = (1/2) * m * v^2.
Rearranging the equation and substituting the given values:
v^2 ≈ 2 * g * h,
v^2 ≈ 2 * 9.8 m/s^2 * 15 m,
v^2 ≈ 294 m^2/s^2.
Therefore, the velocity of the body as it hits the ground is approximately 17.14 m/s.
(c) since the height from which the second body is dropped is the same as the height from which the first body fell, the velocity of the second body just before it hits the ground will also be the same. So, the velocity of the second body as it hits the ground will also be approximately 17.14 m/s.
(d) The mechanical principles applied in the above calculations are the work-energy principle (a) and the conservation of mechanical energy (b, c). The work-energy principle states that the work done on an object equals the change in its kinetic energy. The conservation of mechanical energy states that the total mechanical energy (potential energy + kinetic energy) of a system remains constant if no external forces (apart from gravity) are acting on it.