A 80 kg skier grips a moving rope that is powered by an engine and is pulled at constant speed to the top of a 25 degrees hill. The skier is pulled a distance = 220 m along the incline and it takes 2.3min to reach the top of the hill.
If the coefficient of kinetic friction between the snow and skis is = 0.15, what horsepower engine is required if 30 such skiers (max) are on the rope at one time?

Respuesta :

Answer:

[tex]P=28.085\,hp[/tex]

Explanation:

Given that:

  • mass of 1 skier, [tex]m=80kg[/tex]
  • inclination of hill, [tex]\theta=25^{\circ}[/tex]
  • length of inclined slope, [tex]l=220m[/tex]
  • time taken to reach the top of hill, [tex]t=2.3 min= 138 s[/tex]
  • coefficient of friction, [tex]\mu=0.15[/tex]

Now, force normal to the inclined plane:

[tex]F_N=m.g.cos\theta[/tex]

[tex]F_N=80\times 9.8\times cos25^{\circ}[/tex]

[tex]F_N=710.54\,N[/tex]

Frictional force:

[tex]f=\mu.F_N[/tex]

[tex]f=0.15\times 710.54[/tex]

[tex]f=106.58\,N[/tex]

The component of weight along the inclined plane:

[tex]W_l=m.g.sin\theta[/tex]

[tex]W_l=80\times 9.8\times sin25^{\circ}[/tex]

[tex]W_l=331.33\,N[/tex]

Now the total force required along the inclination to move at the top of hill:

[tex]F=f+W_l[/tex]

[tex]F=106.58+331.33[/tex]

[tex]F=437.91\,N[/tex]

Hence the work done:

[tex]W=F.l[/tex]

[tex]W=437.91\times 220[/tex]

[tex]W=96340.80\,J[/tex]

Now power:

[tex]P=\frac{W}{t}[/tex]

[tex]P=\frac{96340.80}{138}[/tex]

[tex]P=698.12\,W[/tex]

So, power required for 30 such bodies:

[tex]P=30\times 698.12[/tex]

[tex]P=20943.65\,W[/tex]

[tex]P=\frac{20943.65}{745.7}[/tex]

[tex]P=28.085\,hp[/tex]

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