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A person on a sled coasts down a hill and then goes over a slight rise with speed 2.7 m/s. The top of the rise can be modeled as a circle with a radius 4.1 m. The sled and occupant have a combined mass of 110 kg. If the coefficient of friction between the snow and the sled is 0.10, what friction force is exerted on the sled by the snow as the sled goes over the top of the rise?
It would be super helpful if you could explain how/why you came to your answer as well. Thanks!

Respuesta :

egenriether
Since we're dealing with radial acceleration around a circle, I used the radial acceleration equation a=v²/r. At the top of the hill, the force upward exerted by the hill is less than the weight of the sled.  if v is large enough the term (g-v²/r) will become 0 and the sled will fly off the ground as it reaches the peak.  Let me know if I can clarify any of my work.
Ver imagen egenriether

The friction force is exerted on the sled by the snow as the sled goes over the top of the rise is  88.2 N.

Computation:

The force equation for sled+Occupant can be written as :

[tex]Mg-f_n=Ma[/tex]

Since here acceleration is the centripetal acceleration due to motion in a circle

So,

[tex]Mg-f_n=M\frac{V^2}{r}[/tex]

And M = 110 kg , g = 9.8 m/s2 , r = 4.1 m , V = 2.7 m/s then from the above formula,

[tex]f_n=\frac{(110\times 9.5 )-(110\times 2.7}{4.1} \\=882.2N[/tex]

The Formula for frictional force is,

[tex]F_f=u_kF_n\\=0.1\times 882.2\\=88.2N[/tex]

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