A 40 kg-skier starts at the top of a 12-meter high slope. At the bottom, she is traveling 10 m/s. How much energy does she lose
2 answers:
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Answer:
v = √[gR (sin θ - μcos θ)]
Explanation:
The free body diagram for the car is presented in the attached image to this answer.
The forces acting on the car include the weight of the car, the normal reaction of the plane on the car, the frictional force on the car and the net force on the car which is the centripetal force on the car keeping it in circular motion without slipping.
Resolving the weight into the axis parallel and perpendicular to the inclined plane,
N = mg cos θ
And the component parallel to the inclined plane that slides the body down the plane at rest = mg sin θ
Frictional force = Fr = μN = μmg cos θ
Centripetal force responsible for keeping the car in circular motion = (mv²/R)
So, a force balance in the plane parallel to the inclined plane shows that
Centripetal force = (mg sin θ - Fr) (since the car slides down the plane at rest, (mg sin θ) is greater than the frictional force)
(mv²/R) = (mg sin θ - μmg cos θ)
v² = R(g sin θ - μg cos θ)
v² = gR (sin θ - μcos θ)
v = √[gR (sin θ - μcos θ)]
Hope this Helps!!!
