Discover the Relationship between Kinetic Energy and Velocity - FAQs Answered

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Doubling the kinetic energy of an object does not result in a doubling of velocity; instead, it increases velocity by a factor of the square root of two, assuming mass remains constant. The relationship between kinetic energy (KE) and velocity (v) is expressed mathematically as KE = 1/2 mv², indicating that velocity is proportional to the square root of kinetic energy when mass is constant. Discussions also highlight the importance of understanding the work-energy principle, where work done on an object translates to kinetic energy. The force applied over a distance determines the kinetic energy gained, emphasizing that energy is proportional to the distance rather than the time of force application. This understanding clarifies the relationship between kinetic energy and velocity in physics.
Damien20
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If the kinetic Energy of an object is doubled, by what factor is the velocity increased ?
 
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Damien20 said:
If the kinetic Energy of an object is doubled, by what factor is the velocity increased ?
You have to show your work first. Can you tell us how you think velocity is related to kinetic energy?

AM
 
Well I know if the mass constant, and if the velocity is increased or decreased then so does the KE. I can wrap my head around that much, but If the KE is doubled it makes me want to say the velocity and mass doubles. what you do to one side of the equation you must do to the otherside, but for some reason that just doesn't feel right.
 
Damien20 said:
Well I know if the mass constant, and if the velocity is increased or decreased then so does the KE. I can wrap my head around that much, but If the KE is doubled it makes me want to say the velocity and mass doubles. what you do to one side of the equation you must do to the otherside, but for some reason that just doesn't feel right.
You have to give us a mathematical expression for kinetic energy.

If I apply a force to a stationary mass, m, such that F = ma = m\Delta v/\Delta t for a time \Delta t what is the final speed? (ie. what is \Delta v?) How far a distance have I applied that force in that time? (call that d1). The force times that distance d1 is the work done and that is equal to the kinetic energy of the mass (assuming no friction). The energy is proportional to the distance over which I apply the force, not the time over which I apply it. Now if I apply the force for a longer time such that \Delta v is twice as large, over what distance do I apply the force? (d2). The ratio of d2 to d1 gives you the ratio of kinetic energies.

AM
 
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