A runner whose mass is 51 kg accelerates from a stop to a speed of 7 m/s in 2 se

Question

A runner whose mass is 51 kg accelerates from a stop to a speed of 7 m/s in 2 seconds. (A good sprinter can run 100 meters in about 10 seconds, with an average speed of 10 m/s.)

(a) What is the average horizontal component of the force that the ground exerts on the runner's shoes?
Average force =  N

(b) How much displacement is there of the force that acts on the sole of the runner's shoes, assuming that there is no slipping?
Displacement =  m

(c) Therefore, how much work is done on the extended system (the runner) by the force you calculated in part (b)?
Work =  J

(d) How much work is done on the point-particle system by this force? (Hint: use a fundamental principle, as applied to the point-particle system.)
Work =  J

(e) The kinetic energy of the runner increases; what kind of energy decreases? (In this short run, there has not been a significant temperature change in the runner.)

chemical energy

gravitational energy    

translational kinetic energy

thermal energy

rotational kinetic energy

(f) What is the change of the energy you identified in part (e)? Pay attention to signs.
Energy change =  J

Explanation / Answer

a) average F= m(v/t) = 51(7/2) = 178.5N

b) displacement = 0 m

c) Work = 0J

d) Work for the point particle system is: transitional KE = work (extended system)
Because the work (extended system) is 0, we have transitional KE = 1/2mv^2

So, KE=(1/2)(51)(7^2) = 1249.5 J, which is the work

e) Chemical energy

f) it's the opposite of part d, so its -1249.5J

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