Question

A mass m = 90 kg slides on a frictionless track that has a drop, follow by a loop-the-loop with radius R = 15.1 m and at last a flat straight section at the same height as the center of the loop (15.1 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. (Presume the mass never leaves the smooth track at any point on its path.)

1) What is the least speed the block must have at the top of the loop to make it around the loop-the-loop without leaving the track?

2) What height above the ground has got to the mass begins to make it around the loop-the-loop?

3) If the mass has just sufficient speed to make it around the loop without leaving the track, what will its speed be at the bottom of the loop?

4) If the mass has just enough speed to make it around the loop without leaving the track, what is its speed at final flat level (15.1 m off the ground)?

5) Now a spring with spring constant k = 17400 N/m is used on the final flat surface to stop the mass. How far does the spring condense?

6 ) It turns out the engineers designing the loop-the-loop didn't really know physics - as they made the ride, the first drop was only as high as the top of the loop-the-loop. To account for the mistake, they decided to give the mass an initial velocity right at the beginning. How fast do they require to push the mass at the beginning (now at a height equal to the top of the loop-the-loop) to get the mass around the loop-the-loop without falling off the track?

7) The work done by the normal force on the mass (during the initial fall) is:
a. positive
b. zero
c. Negative