The picture below shows a cart at the top of a hill of height h_1 = 15.0 m. The

Question

The picture below shows a cart at the top of a hill of height h_1 = 15.0 m. The carts from rest and rolls down to the bottom of the first hill and then over the top of a second hill of height h_2 = 12.5 m. The top of the second hill is round with a radius of curvature R = 6.25 m. The combined mass of the cart and its rider is 450.0 kg. The effect of friction on the cart as its rolls can be neglected. a. How fast is the cart moving when it reaches the top of the second hill? b. What is the normal force on the cart as it moves over the top of the second hill?

Explanation / Answer

Given

mass of the cart and rider is m = 450 kg

h1 = 15 m, h2 = 12.5 m

R = 6.25 m

a) conservation of energy

gravitaional potential energy = kinetic energy

   mgh1 = 0.5*mv^2

   v^2 = 2gh1

   v = sqrt(2gh1)

   v = sqrt(2*9.8*15) m/s

   v = 17.1464282 m/s

the kinetic energy at ground is = 0.5*mv^2 = 0.5*450*17.1464282^2 J = 66150 J

when it reaches the top of the second hill the speed is

   mgh2 = 0.5*mv^2

   v = sqrt(2gh2)
   v = sqrt(2*9.8*12.5) m/s

   v = 15.6524758425 m/s

b) the normal force is N = mg = 450*9.8 N = 4410 N

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