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| Illustration of Conservation of Mechanical Energy |
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 Conversion of gravitational potential energy to kinetic energy: When a body falls from a certain height, its kinetic energy increases. |
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| The gain in kinetic energy is at the expense of the gravitational potential energy. But the total mechanical energy remains conserved. |
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| Let us take the case of a freely falling body. Consider a body of mass 'm' to be at a position A. Let 'h' be the height of the body above the reference level. |
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| At the position A |
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| Kinetic energy of the body = 0 |
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| (Since the body is at rest, i.e., v = 0) |
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| Potential energy of the body = mgh. |
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| Therefore the total mechanical energy = mgh + 0 = mgh |
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| At position B |
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| Let the body be at the position B at any instant, after having fallen through a distance x. |
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| Potential energy of the body = mg (h - x) |
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| If vB is the velocity of the body at B, |
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| then vB2- 02 = 2gx |
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 vB2 = 2gx |
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| Total energy of the body = mg (h - x) + mg x = mgh |
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| At the position C |
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| When the body is at position C, its height above the ground may be regarded as zero. Therefore, potential energy of the body = 0. If vc is the velocity with which the body touches the ground, then vc2- 0 = 2gh |
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 vc2 = 2gh |
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| Total energy of the body = mgh |
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| From the above illustration, it is clear that the sum of the kinetic and potential energies of a freely falling body is constant at all stages of motion. As the body falls, its potential energy decreases while its kinetic energy increases, but the total mechanical energy is conserved. |
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| The graph depicts the value of kinetic energy and potential energy at different heights above the reference level. |
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| The potential energy decreases linearly with decrease in height. On the other hand, the kinetic energy increases with decrease in height. The horizontal line represents the total mechanical energy. |
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| Let us now consider a body sliding down a smooth inclined plane. Consider a smooth inclined plane of length 'l' and height 'h' and 'q' be the angle of inclination. Let a body of mass 'm' be at rest at the top (A) of the inclined plane. |
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| Potential energy of the body = mgh |
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| Kinetic energy of the body = 0 |
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| Total energy = mgh |
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| Let vx be the velocity acquired by the body in sliding through a distance x. |
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| Then vx2- 02 = 2ax but a = g sinq |
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| Potential energy = mgh2 |
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| Total energy = mgh1 + mgh2 |
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| = mg (h1 + h2) |
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| = mgh |
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| Let vl be the velocity of the body at the bottom B of the inclined plane. |
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| Then vl2 - 02 = 2 x g sinx l |
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= 2gl sin |
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| Potential energy at B = 0 |
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 Total energy = mgh |
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| The above equations show that the total mechanical energy is conserved. |
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| The figure shows a vibrating simple pendulum, O is the point of suspension. When the centre of gravity of the bob is vertically below the point of suspension, the position of the pendulum is called the rest position or mean position. |
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| Let us say that the bob is displaced to a position E1. When released, the pendulum will start vibrating between the two extreme positions E1 and E2. |
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| When the bob moves from mean position, to the extreme position E1, its centre of gravity is raised vertically through a height 'h'. |
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 Potential energy of the bob at the extreme position = mgh |
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| As the bob moves from the extreme position to the mean position, its potential energy goes on decreasing while the kinetic energy goes on increasing. But, the sum of the two energies is constant. At the mean position, the whole of the potential energy is converted into kinetic energy. |
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| If 'v' is the velocity of the bob at the mean position, then |
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| Applying the law of conservation of energy, |
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| As the bob moves from mean position to the other extreme position E2, its kinetic energy decreases and potential energy increases. Energy becomes wholly potential at E2. As the bob moves from E2 to the mean position, the potential energy goes on decreasing and kinetic energy goes on increasing. At the mean position, the energy is once again, wholly kinetic. |
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