High School Is Work the Same as Kinetic Energy?

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Kinetic Energy (KE) is derived from the formula KE = ½(m⋅v²), which can be expressed in terms of force and displacement as KE = F⋅s, where F is force and s is displacement. This relationship leads to the conclusion that work done on an object results in an increase in its kinetic energy, a principle known as the Work-Energy Theorem. The discussion highlights that this theorem applies under conditions of straight-line motion, constant force, and constant acceleration, but it is a broader concept applicable in various scenarios. For example, when throwing a ball, its height can be calculated by equating its initial kinetic energy to the work done against gravity. Overall, the connection between work and kinetic energy emphasizes the conservation of energy in physical systems.
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Kinetic Energy is defined as the product of half the mass of an object multiplied by the square of its velocity. Using this I derived the end formula. Work is as follows:

KE = ½(m⋅v2)

Because v = vi +at, if vi = 0,

KE = ½(m⋅(at)2)

KE = ½(m⋅a2⋅t2)

KE = ½(m⋅a⋅a⋅t2)

Substituting F for m⋅a (force is equal to mass by acceleration),

KE = ½(F⋅a⋅t2)

Displacement is equivalent to vi⋅t +½⋅a⋅t2
It was stated that vi = 0, therefore s = ½⋅a⋅t2
Making this replacement in the equation yields:

KE = F⋅ s

Where F is a force and s is a displacement. This is also the formula for work. Then, is work the same thing as Kinetic Energy? Can the formula for work be used to calculate Kinetic Energy?
 
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Yes, that's often called the "Work-Energy Theorem". It's just conservation of energy. If you put work into the motion of an object, that object's kinetic energy increases by the amount of work you put in.

Your derivation used some formulas that assume straight line motion, constant force and constant acceleration. But it's a more general result than that. If you put energy into an object, the object's energy increases by that much.

That concept can be used in all sorts of situations. Suppose you throw a ball up in the air. How high will it go? Answer: It will rise until its initial kinetic energy (1/2)mv^2 has all been turned into work against gravity mg * h. So you set those two things equal and solve for h. KE = work.

If you have an object moving with velocity v and a frictional force F acts on it, how far does it slide before friction stops it? It takes until the work Fd done by friction is equal to the amount of energy (1/2)mv^2 that the object started with. Again, set KE = work.
 
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I do not have a good working knowledge of physics yet. I tried to piece this together but after researching this, I couldn’t figure out the correct laws of physics to combine to develop a formula to answer this question. Ex. 1 - A moving object impacts a static object at a constant velocity. Ex. 2 - A moving object impacts a static object at the same velocity but is accelerating at the moment of impact. Assuming the mass of the objects is the same and the velocity at the moment of impact...

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