How much work is needed for a 67 kg runner to accelerate from rest to 10.5 m/s?

Question

in kJ

Answer 1

well, given that work is energy, and kinetic energy is K=1/2(mv^2)

K1(starting kinetic energy)=0 (since v=0)
K2(kinetic energy at v=10.5)=1/2(67x10.5^2)=3693.375j=3.693375Kj

so the difference in energy will tell you how much is needed:
work done = 3.693375-0 Kj
work done = 3.693375 Kj

Answer 2

This is a conservation of energy problem. The runner's total energy before starting is TE(0) = PE = potential energy because there is no kinetic energy, KE = 0, when standing still. The PE in this instance is the runner's electro-chemical energy in the muscles that make the runner move and accelerate.

Upon starting, we have TE(v > 0) = PE + WE + KE; where WE = FS; where F = ma is the runner's motivating force, m is 67 kg, a is the runner's acceleration, and S is the distance the runner covers with that force.

The work energy is put in by the runner to convert her PE into KE the kinetic energy. That is, as work is done potential energy diminishes a bit and KE increases a like amount. Total energy TE remains the same however because of the conservation of energy.

When the runner reaches V, TE(v = 10.5) = PE + KE; there is still PE because there is still potential muscle energy. But now v = 10.5; so there is kintetic energy as well.

From the conservation of energy we can equate the total energies. Thus, TE(v>0) = PE + WE = PE + maS = PE + 1/2 mv^2 = PE + KE = TE(v = 10.5). In which case, WE = maS = 1/2 mv^2; so the "work needed" is WE = 1/2 mv^2. You have m and v given; so you can do the math.

Now the physics...The PE's cancel out because muscle energy is available both before starting and when v = 10.5 mps. And, reason tells us, the electro-chemical potential energy of the runner will be about the same in both instances. What has happened, with the work is that some PE was converted into KE, and that's where v > 0 came from.