How Is Work Calculated in a Point-Particle System for a Sprinting Runner?

[lebprince]

Homework Statement

A runner whose mass is 49 kg accelerates from a stop to a speed of 6 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? I got the answer for this which is 147 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? I got the answer for this one too = 0 m

(c) Therefore, how much work is done on the real system (the runner) by the force you calculated in part (b)? I also got 0 for this one since its F x d

(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.)

The Attempt at a Solution

I got a,b,c am just having problems with part d. Thanks a lot

 

[lebprince]

Homework Statement

A runner whose mass is 49 kg accelerates from a stop to a speed of 6 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? I got the answer for this which is 147 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? I got the answer for this one too = 0 m

(c) Therefore, how much work is done on the real system (the runner) by the force you calculated in part (b)? I also got 0 for this one since its F x d

(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.)

The Attempt at a Solution

I got a,b,c am just having problems with part d. Thanks a lot

figured it out thanks

 

[thomate1]

for the help!


I would like to commend you on your accurate calculations for parts (a), (b), and (c). For part (d), we can use the fundamental principle of work and energy, which states that the work done on a system is equal to the change in its kinetic energy. In this case, the point-particle system refers to the runner's entire body, which can be approximated as a single particle with a mass of 49 kg.

Since the runner starts from rest and accelerates to a speed of 6 m/s, the change in kinetic energy is given by:

ΔKE = ½ * m * v^2 = ½ * 49 kg * (6 m/s)^2 = 882 J

Therefore, the work done on the point-particle system by the ground's force is equal to 882 J. This is the same as the work done on the real system (the runner) in part (c), as the two systems are equivalent in this case. I hope this helps to answer your question. Keep up the good work in your studies!