A 61.0-kg runner has a speed of 3.20 m/s at one instant during a long-distance event.
(a) What is the runner's kinetic energy at this instant?
KEi = J
(b) If he doubles his speed to reach the finish line, by what factor does his kinetic energy change?
KEf
KEi
=
2 answers
Think I answered this once
KE = 1/2 mv^2, so just plug in your numbers.
Now, if we have 2v instead of v, that gives us
1/2 m(2v)^2 = 1/2 m*4v^2 = 4(1/2 mv^2)
since KE varies as v^2, 2v gives 2^2 times the energy.
Now, if we have 2v instead of v, that gives us
1/2 m(2v)^2 = 1/2 m*4v^2 = 4(1/2 mv^2)
since KE varies as v^2, 2v gives 2^2 times the energy.
1 answer