Please help me in tge circled question
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ormula for (K.E is 1/2mv^2)_eq 1
So if increased m 4 time we should decrease it velocity by half
By increasing mass by four time eq 1 become
[1/2 (4m)(v)^2]_eq 2
To maintain same kinetic energy decrease it velocity by half
eq 2 become
》1/2 (4m)(v/2)^2
》(4m×v^2/4)/2
》1/2mv^2
What change should be expected in the velocity of a body to maintain the same kinetic energy, if its mass is increased sixteen times? How?
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BORYS SHUMYATSKIY eNotes educator| CERTIFIED EDUCATOR
Hello!
Kinetic energy is that part of full energy which a body has due to its motion. The formula for kinetic energy is E_k = (m V^2)/2, where m is the mass and V is the speed (regardless of direction).
Usually a body remains the same during its motion, and the mass of the body also remains the same. In our problem, the mass of the body is supposed to increase 16 times, roughly speaking...
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Hello!
Kinetic energy is that part of full energy which a body has due to its motion. The formula for kinetic energy is E_k = (m V^2)/2, where m is the mass and V is the speed (regardless of direction).
Usually a body remains the same during its motion, and the mass of the body also remains the same. In our problem, the mass of the body is supposed to increase 16 times, roughly speaking some other bodies will join our initial body.
In such a case, its kinetic energy becomes E'_k = ((16 m) V^2)/2 = 16 E_k. To compensate this change by a speed change, we have to reduce V^2 16 times, which means to reduce V sqrt(16)=4 times.
This is the answer: body's speed must be reduced 4 times to maintain the same kinetic energy.
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