Physics - acceleration?

As the runner moves around the curve he must exert enough force to change the direction of his velocity by 45˚. This is one eighth of a circle. Use the following equation to determine the force.

Fc = m * v^2/r = 65 * 4.4444^2/5 ≈ 256.7 N

This is the force which he must exert to change the direction of his velocity by 45˚.

His acceleration is v^2/r = 4.4444^2/5 ≈ 3.95 m/s^2

The direction of the acceleration is toward the center of the circle. This acceleration causes his angular velocity to increase.

Rotational KE = ½ * I * ω^2

I = m * r^2 = 65 * 5^2 = 4225

ω = v/r = 4.4444/5 = 0.8888 rad/s

Rotational KE = ½ * 4225 * 0.8888^2 ≈ 1668.8 J

This is the additional energy required to run one 45˚ curve

1b. The runner does another mile, everything is all even and equal except this time the radius of the corner is 10m. How much additional energy will he use to turn this corner? In both cases his total distance is 1 mile.

I = m * r^2 = 65 * 10^2 = 6500

ω = v/r = 4.4444/5 = 0.8888 rad/s

Rotational KE = ½ * 6500 * 0.8888^2 ≈ 2567.4 J

This is the additional energy required to run one 45˚ curve

2. The runner now runs a straight mile but climbs 10m over the distance - how much energy does he use to gain this height (Just to confirm for me that this is mass x height gained)

Yes, energy = 65 * 9.8 * 10 = 6370 J

I agree with your first answer. He didn't comment on your 1b -- probably since the work is still zero. But let me point out that a 10 m radius is a more gentle curve so the centripetal force would be less than for the 5 m radius. He also didn't point out that the "x g" was missing in your comment "Just to confirm for me that this is mass x height gained". The units of the product of mass and height are wrong to be energy.

Regarding your additional details:

energy = force x distance x cosine of the angle between them. That angle is 90 degrees in this case and cos90=0. The first answer touched on that.

I might allow you to quibble on one point. The energy consumed or transferred by a human body is more complicated than if we were talking about a machine. A man holding weights in his hands at waist height for 60 seconds has to feed energy to the muscles in his arms to keep them up there. If he puts the weights on a shelf at waist height, the shelf does not do any work holding the weights. I'm burning energy as I sit here but I'm not doing any work in a physics sense (ignoring a small amount involved with pushing buttons). This is a physics site, so the first answer answered in that sense.

EDIT: I added 1a, 1b et cetera because although I answered all, another answerer did not spot it without these placeholders...

Work equals force times displacement IN THE DIRECTION OF THE FORCE.

1a) Because the (centripetal) force is by definition perpendicular to the displacement, there is no additional work done. Incidentally, you might have seen this from the fact that kinetic energy isn't changeing in making the turn, so the work-energy theorem would have allowed you to conclude that the work is zero.

1b)The radius of the circular bend does not matter, the force remains perpendicular to the displacement.

1c) For the climb, the force to get up is vertical and equals mg, and the vertical displacement is h, so the amount of work is

W = m g h

Note that this work will be "stored energy" we call gravitational potential energy U. That's why it is

U = m g h.