The Energy and Efficiency of Cross-Country Skiing: A Mathematical Analysis

[snav96]

1. A cross-country skier ascends a peak and rises 300 m vertically during the ascent. Her mass including clothing skis, poles, shoes and backpack is 65 kg.
   
   1. (a) Find the minimum mechanical work needed to be done by the skier (neglect friction).
   2. (b) Why do we not need to consider the skier’s kinetic energy in part (a)?
   3. (c) What is the metabolic rate needed, if the efficiency for this activity is 11% and it took 30 min to climb up that 300 m? (‘rate’ here means energy/time, i.e. power).
   4. (d) If the mean sliding friction force opposing motion of the skis when slid forward with each stride is 4.5 N, how much energy is dissipated by friction in a distance of 1200 m measured along the path?
      

Homework Equations

The Attempt at a Solution

a) Potential Energy = 300 m x 9.8 x 65 Kg = 191100 J

b) because the speed stays constant. there's no change in kinetic energy [/B]

c) amount of efficient energy = 191100 x 0.11 = 21021 J
Power = 21021 / 30 x 60 = 11.68 w

d) ?

can anyone check to see if my answers are correct? and help me with part d? cause i have no idea how to answer part d

thanks

 

[vela]

A cross-country skier ascends a peak and rises 300 m vertically during the ascent. Her mass including clothing skis, poles, shoes and backpack is 65 kg.

1. (a) Find the minimum mechanical work needed to be done by the skier (neglect friction).
2. (b) Why do we not need to consider the skier’s kinetic energy in part (a)?
3. (c) What is the metabolic rate needed, if the efficiency for this activity is 11% and it took 30 min to climb up that 300 m? (‘rate’ here means energy/time, i.e. power).
4. (d) If the mean sliding friction force opposing motion of the skis when slid forward with each stride is 4.5 N, how much energy is dissipated by friction in a distance of 1200 m measured along the path?

Homework Equations

The Attempt at a Solution

[/b]

a) Potential Energy = 300 m x 9.8 x 65 Kg = 191100 J

b) because the speed stays constant. there's no change in kinetic energy

c) amount of efficient energy = 191100 x 0.11 = 21021 J
Power = 21021 / 30 x 60 = 11.68 w

Efficiency is (useful output)/input. The useful output in this problem is the work needed to ascend the 300 m.

d) ?

Can anyone check to see if my answers are correct, and help me with part d Because I have no idea how to answer part d?

Part d is asking you to calculate the work done by friction.

 

[snav96]

Efficiency is (useful output)/input. The useful output in this problem is the work needed to ascend the 300 m.Part d is asking you to calculate the work done by friction.

so it is it just 4.5N x 1200m ?

 

[vela]

Yup.