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solong [7]
3 years ago
5

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!

Physics
1 answer:
Paladinen [302]3 years ago
8 0
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.

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A 90-kg astronaut is stranded in space at a point 6.0 m from his spaceship, and he needs to get back in 4.0 min to control the s
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v = 2/90 

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8 0
3 years ago
A roller-coaster car has a mass of 1080 kg when fully loaded with passengers. As the car passes over the top of a circular hill
Dmitriy789 [7]

Answer:

(a): The normal force on the car  from the track when the car's speed is v= 7.6 m/s  is  FN= -6696 N.

(b): The normal force on the car from the track when the car's speed is v= 17 m/s is FN= 8912.7 N.

Explanation:

m= 1080 kg

r= 16m

v1= 7.6 m/s

v2= 17 m/s

g= 9.81 m/s²

v1= w1*r

w1= v1/r

w1= 0.475 rad/s

ac1= w1² * r

ac1= 3.61 m/s²

FN= m * (ac1 - g)

FN= -6696 N    (a)

-----------------------------------------------------

v2= w2*r

w2= v2/r

w2= 1.06 rad/s

ac2= w2² * r

ac2= 18.06 m/s²

FN= m * (ac2 - g)

FN= 8912.7 N    (b)

4 0
3 years ago
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