which mathematical relaction of energy of a stone of mass m Falling freely from hight h remains conserved at every point in its downward motion?
Answers
Explanation:
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pls find the answer below
Law of conservation of energy states that the energy can neither be created nor destroyed but can be transformed from one form to another.
Let us now prove that the above law holds good in the case of a freely falling body.
Let a body of mass 'm' placed at a height 'h' above the ground, start falling down from rest.
In this case we have to show that the total energy (potential energy + kinetic energy) of the body at A, B and C remains constant i.e, potential energy is completely transformed into kinetic energy.
At A,
Potential energy = mgh
Kinetic energy = 1/2 mv² = 1/2 * m* 0Kinetic energy = 0 [the velocity is zero as the object is initially at rest]
\ Total energy at A = Potential energy + Kinetic energy
= mgh + 0
Total energy at A = mgh …(1)
At B,
Potential energy = mgh
= mg(h - x) [Height from the ground is (h - x)]Potential energy = mgh - mgx
Kinetic energy = 1/2 mv²
The body covers the distance x with a velocity v. We make use of the third equation of motion to obtain velocity of the body.
v2 - u2 = 2aSHere, u = 0, a = g and S = x
v² - 0 = 2gx
Here, u = 0, a = g and S = x
v² - 0 = 2gx
v² = 2gx
kinetic energy = 1/2 mv²
= 1/2 m2gx
Kinetic energy = mgx
Total energy at B = Potential energy + Kinetic energy
= mgh - mgx + mgx
Total energy at B = mgh …(2)
At C,
Potential energy = m x g x 0 (h = 0)
Potential energy = 0Kinetic energy = 1/2 mv²
The distance covered by the body is hv2 - u2 = 2aS
Here, u = 0, a = g and S = h
v² - 0 = 2gh
v² = 2gh
kinetic energy = 1/2 mv²
= 1/2 m * 2gh
Kinetic energy = mgh
Total energy at C = Potential energy + Kinetic energy= 0 + mgh
Total energy at C = mgh …(3)It is clear from equations 1, 2 and 3 that the total energy of the body remains constant at every point. Thus, we conclude that law of conservation of energy holds good in the case of a freely falling body.
pls find the answer below
Law of conservation of energy states that the energy can neither be created nor destroyed but can be transformed from one form to another.
Let us now prove that the above law holds good in the case of a freely falling body.
Let a body of mass 'm' placed at a height 'h' above the ground, start falling down from rest.
In this case we have to show that the total energy (potential energy + kinetic energy) of the body at A, B and C remains constant i.e, potential energy is completely transformed into kinetic energy.
At A,
Potential energy = mgh
Kinetic energy = 1/2 mv² = 1/2 * m* 0Kinetic energy = 0 [the velocity is zero as the object is initially at rest]
\ Total energy at A = Potential energy + Kinetic energy
= mgh + 0
Total energy at A = mgh …(1)
At B,
Potential energy = mgh
= mg(h - x) [Height from the ground is (h - x)]Potential energy = mgh - mgx
Kinetic energy = 1/2 mv²
The body covers the distance x with a velocity v. We make use of the third equation of motion to obtain velocity of the body.
v2 - u2 = 2aSHere, u = 0, a = g and S = x
v² - 0 = 2gx
Here, u = 0, a = g and S = x
v² - 0 = 2gx
v² = 2gx
kinetic energy = 1/2 mv²
= 1/2 m2gx
Kinetic energy = mgx
Total energy at B = Potential energy + Kinetic energy
= mgh - mgx + mgx
Total energy at B = mgh …(2)
At C,
Potential energy = m x g x 0 (h = 0)
Potential energy = 0Kinetic energy = 1/2 mv²
The distance covered by the body is hv2 - u2 = 2aS
Here, u = 0, a = g and S = h
v² - 0 = 2gh
v² = 2gh
kinetic energy = 1/2 mv²
= 1/2 m * 2gh
Kinetic energy = mgh
Total energy at C = Potential energy + Kinetic energy= 0 + mgh
Total energy at C = mgh …(3)It is clear from equations 1, 2 and 3 that the total energy of the body remains constant at every point. Thus, we conclude that law of conservation of energy holds good in the case of a freely falling body.