a) Demon Drop Vertical (The drop part): Suppose the cage full of people of mass

Question

a) Demon Drop Vertical (The drop part): Suppose the cage full of people of mass 1.25x103 kg falls under the influence of the gravitational force for a total vertical distance of 30.2 m without experiencing any friction forces (this is an idealization!). How much work does the gravitational force do on the cage? Use the work-energy theorem to calculate the kinetic energy of the cart at the point where it hits the horizontal part of the track b)Demon Drop Slow Down (The stopping part): If the brakes are applied just as the cart hits the level part of the track and the cart stops in a distance of 21.0 m, use the work-energy theorem to find the coefficient of friction between the brake pads and the horizontal portion of the track. Physics

Explanation / Answer

(a)

at the beginning the cage jas potential energy which is transfered to the kinetic energy;

mhg=mv^2/2

work done by gravity is F*d=mg*d=1.25e3*9.8*30.2=3.7e5 J

kinetic energy = mv^2/2=mgh=1.25e3*9.8*30.2 = 3.7e5 J

(b)

friction force does the work through 21 meters and it brings the cart to the rest

F=umg

F*d=Ek --> umg*21 = 3.7e5--> u=1.438

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