A tennis ball has a mass of 0.057 kg. How much force will need to be exerted to make the ball accelerate at a rate of 570? force:(0.057*570)= 32.5 N
mass:0.057 kg
acceleration:570 m/s^2

Amass m = 74 kg slides on a frictionless track that has a drop, followed by a loop-the-loop with radius r = 18.4 m and finally a flat straight section at the same height as the center of the loop (18.4 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? 2. what height above the ground must the mass begin to make it around the loop-the-loop? 3. if the mass has just enough 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 the final flat level (18.4 m off the ground)? 5. now a spring with spring constant k = 15600 n/m is used on the final flat surface to stop the mass. how far does the spring compress?