You are in charge of testing/designing a new amusement park ride. At the end of

Question

You are in charge of testing/designing a new amusement park ride. At the end of the ride, the 1200 kg cart (including people) goes down a hill that is angled at 30o from the horizontal. The starting height of the cart is 20 m above the ground, and it is going 5 m/s at the top of the hill. There is a constant force of friction of 30 N on the cart the entire ride. At the bottom of the hill, the cart travels 50 m before being stopped by a spring that is compressed by 2 m. What is the spring constant of the spring?

Explanation / Answer

the kinetic energy of the cart at the hill top is

K = (1/2)mv^2

where m = 1200 kg and v = 5 m/s

Also,W = F x S

where F is force and S is distance(20 m)

or F x S = (1/2)mv^2

or F = (1/2S)mv^2

the net force acting on the cart is

F_net = F - f_k

where f_k is force of friction(30 N)

the kinetic energy of the cart before coming to stop is

K1 = F_net x S1

where S1 = 50 m

the above kinetic energy is equal to the kinetic energy of the spring therefore

K1 = (1/2)k x d^2

or F_net x S = (1/2)k x d^2

or k = (2F_net x S/d^2)

where k is spring constant of spring and d = 2 m

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