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A 64.0-kg skier starts from rest at the top of a ski slope of height 62.0 m.
A)If frictional forces do -1.10×104 J of work on her as she descends, how fast is she going at the bottom of the slope?
Take free fall acceleration to be g = 9.80 m/s^2.
B)Now moving horizontally, the skier crosses a patch of soft snow, where the coefficient of friction is 0.18. If the patch is of width 70.0 m and the average force of air resistance on the skier is 160 N, how fast is she going after crossing the patch?
C)After crossing the patch of soft snow, the skier hits a snowdrift and penetrates a distance 3.0 m into it before coming to a stop. What is the average force exerted on her by the snowdrift as it stops her?

Answer :

Answer:

Part a)

[tex]v = 29.5 m/s[/tex]

Part b)

[tex]v_f = 16.5 m/s[/tex]

Part c)

[tex]F = 2915 N[/tex]

Explanation:

Part a)

As per work energy theorem we can say that work done by all forces on the ski must be equal to change in kinetic energy

so we will have

[tex]W_g + W_{friction} = \frac{1}{2}m(v_f^2 - v_i^2)[/tex]

now we will have

[tex]mgH - 1.10 \times 10^4 = \frac{1}{2}(64)(v^2 - 0)[/tex]

[tex]64(9.80)(62) - 1.10 \times 10^4 = 32 v^2[/tex]

[tex]v = 29.5 m/s[/tex]

Part b)

Now when the ski mover through a rough path then again we can use work energy theorem to find the final speed after crossing the patch

[tex]W_f + W_{air} = \frac{1}{2}m(v_f^2 - v_i^2)[/tex]

[tex]-F_{air} d - \mu mg d = \frac{1}{2}m(v_f^2 - v_i^2)[/tex]

[tex]-(160)(70) - 0.18(64\times 9.8)(70) = \frac{1}{2}(64)(v_f^2 - 29.5^2)[/tex]

[tex]-19102.72 = 32(v_f^2 - 29.5^2)[/tex]

[tex]v_f = 16.5 m/s[/tex]

Part c)

As the ski strike with snow drift then after moving a distance of d = 3.0 m the ski will stop

so here we can say that work done while it penetrate is equal to change in kinetic energy

[tex] -F. d = \frac{1}{2}m(v_f^2 - v_i^2)[/tex]

[tex]- F. 3 = \frac{1}{2}(64)(0 - 16.5^2)[/tex]

[tex]F = 2915 N[/tex]

okpalawalter8

The speed at bottom of the slope, speed crossing the patch, average force exerted the snowdrift  are mathematically given as

  • V=29.99m/s
  • v=17.395m/s
  • Esnow=3227.58N

The speed at bottom of the slope, speed crossing the patch and average force exerted the snowdrift  

Generally the equation for the conservation of energy  is mathematically given as

[tex]V=\sqrt{2(mgh+Wf)/2}\\\\Therefore\\\\\V=\sqrt{2(mgh+Wf)/2}\\\\V=\sqrt{2(64*9.01*62)-1.10*10^4)/62.0}[/tex]

V=29.99m/s

B)

Efi=1/2mv^2

Efi=1/2*64*29.99^2

Efi=28785.96J

Hence

Eff=Efi-umgd-Fd

0.5*64*v^2=28785-(0.18*64*9.80*70.0)-(160*70)

[tex]v'=\sqrt{302.6}[/tex]

v=17.395m/s

c)

Eff=W

Esnow=0.5mv'^2/d'
Therefore

Esnow=1/2*64*(17.395)^2/3.0

Esnow=3227.58N

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