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You are a member of an alpine rescue team and must project a box of supplies, with mass m, up an incline of constant slope angle a so that it reaches a stranded skier who is a vertical distance h above the bottom of the incline. The incline is slippery, but there is some friction present, with kinetic friction coefficient µk.
Use the work-energy theorem to calculate the minimum speed v that you must give the box at the bottom of the incline so that it will reach the skier.
Express your answer in terms of some or all of the variables m, g, h, µk, and a.

Answer :

Answer:

[tex]v = \sqrt{2gh + \frac{2\mu_kgh}{tan\theta}}[/tex]

Explanation:

When we push the box from the bottom of the incline towards the top then by work energy theorem we can say that

Work done by all the forces = change in kinetic energy of the system

[tex]- mgh - F_f (s) = 0 - \frac{1}{2}mv^2[/tex]

here we know that

[tex]F_f = \mu_k mg cos\theta[/tex]

also we know that the length of the incline is given as

[tex]s = \frac{h}{sin\theta}[/tex]

now we have

[tex]- mgh - \mu_k mgcos\theta(\frac{h}{sin\theta}) = -\frac{1}{2}mv^2[/tex]

so we have

[tex]v = \sqrt{2gh + \frac{2\mu_kgh}{tan\theta}}[/tex]

Cricetus

The expression will be "[tex]v = \sqrt{2gh (1+ \mu \ Cot a)}[/tex]".

Work-energy theorem:

As we know,

→ Kinetic energy loss = Potential energy gain + WD against friction

Now,

  • Initial Kinetic energy = [tex]\frac{1}{2} mv^2[/tex]
  • Potential energy gain = mgh
  • WD against friction = [tex]Ff\times L[/tex]

→ [tex]Ff = \mu\times mg.Cos \ a[/tex]

→ [tex]L = \frac{h}{Sin \ a}[/tex]

By placing the values, we get

→ [tex]WD = \mu.mg.h \ Cot \ a[/tex]

 [tex]\frac{1}{2} mv^2 = mg.h + \mu. mg.h \ Cot \ a[/tex]

        [tex]v = \sqrt{2gh(1+\mu \ Cot \ a)}[/tex]

Thus the above response is right.      

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