In tests on earth a lunar surface exploration vehicle (mass=5.90*103 kg) achieves a forward acceleration of 0.220m/s2. to achieve this same acceleration on themoon, thevehicle's engines must produce a drive force of 1.43*103n. what is the magnitude of the frictional force that acts on thevehicle on the moon? all i know is it has something to do with the formulag=m1m2/r2. i figure the thefrictional force is th force opposing the drive force..i figured it out !

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?