A mass m = 86 kg slides on a frictionless track that has a drop, followed by a l

Question

A mass m = 86 kg slides on a frictionless track that has a drop, followed by a loop-the-loop with radius R = 19.5 m and finally a flat straight section at the same height as the center of the loop (19.5 m off the ground). Since the mass would not make it around the loop if released from the height of the top of the loop (do you know why?) it must be released above the top of the loop-the-loop height. (Assume the mass never leaves the smooth track at any point on its path.

1) a spring with spring constant k = 1.5 × 104 N/m is used on the final flat surface to stop the mass. How far does the spring compress?

Explanation / Answer

Given data ; mass m = 86 kg, radius R = 19.5 m, height h = 19.5 m, K =1.5*10^4 N/m

The height above the ground must the mass begin to make it around the loop-the-loop,

h = (5/2)*R

= 48.75 m

the spring compress is

x = (3*m*g*R/k)^1/2

= sqrt(3*86*9.8*19.5/1.5*10^4)

x = 1.812 m

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