RRA mass m = 88 kg slides on a frictionless track that has a drop, followed bya loop-the-loop with radius R = 15.8 m and finally a flat straight section atthe same height as the center of the loop (15.8 m off the ground). Sincethe mass would not make it around the loop if released from the height ofthe top of the loop (do you know why?) it must be released above the topof the loop-the-loop height. (Assume the mass never leaves the smoothtrack at any point on its path.) 4) If the mass has just enough speed to make it around the loop withoutleaving the track, what is its speed at the final flat level (15.8 m off theground)?m/sSubmitm5) Now a spring with spring constant k = 16500 N/m is used on the final flatsurface to stop the mass. How far does the spring compress?+Submit+6) It turns out the engineers designing the loop-the-loop didn't really knowphysics when they made the ride, the first drop was only as high as thetop of the loop-the-loop. To account for the mistake, they decided to givethe mass an initial velocity right at the beginning.How fast do they need to push the mass at the beginning (now at a heightequal to the top of the loop-the-loop) to get the mass around the loop-the-loop without falling off the track?m/s Submit+

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R R A mass m = 88 kg slides on a frictionless track that has a drop, followed by a loop-the-loop with radius R = 15.8 m and finally a flat straight section at the same height as the center of the loop (15.8 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.)

R R A mass m = 88 kg slides on a frictionless track that has a drop, followed by a loop-the-loop with radius R = 15.8 m and finally a flat straight section at the same height as the center of the loop (15.8 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.)

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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R R A mass m = 88 kg slides on a frictionless track that has a drop, followed by a loop-the-loop with radius R = 15.8 m and finally a flat straight section at the same height as the center of the loop (15.8 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) What is the minimum speed the block must have at the top of the loop to make it around the loop-the-loop without leaving the track? m/s Submit 2) What height above the ground must the mass begin to make it around the loop-the-loop? m Submit 3) If the mass has just enough speed to make around the loop without leaving the track, what will its speed be at the bottom of the loop? m/s Submit +
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