1.) Consider the forces acting at the bottom of the loop:

a. Normal Force points Upwards

b. Gravity Points Downwards

The net force (centripetal force) also points upwards, therefore:

mv^2/r = N - mg

mv^2/r + mg = N

The velocity here is the max velocity at the bottom of the ramp. The radius is the radius of the loop. The mass is the mass of the block. We can plug in the values they provided to find the normal force, but first using conservation of mechanical energy, let's find the final velocity.

Initially we have PE only. At the bottom of the ramp, all that energy is converted entirely to KE. (We're told there's no friction).

mgh = 1/2mv^2

Solving for v, we get: v = sqrt(2gh); v = sqrt(2x10x80); v = sqrt(1600); v = 40m/s

Note: The height of the loop is 2x the radius: 20m. The height of the ramp is 4 times the height of the loop, 80m.

Plugging in this velocity to find Normal Force, we get:

(1kg)(40m/s)^2/(10m) + (1kg)(10m/s^2) = N

160N + 10N = 170N

Normal Force = 170N

2.) Centripetal Acceleration = v^2/r. We can solve for tangetial velocity and use conservation of ME. 4g = 40m/s^2. 40m/s^2 x 10m = 400 = v^2. sqrt(400) = v; v=20m/s

Therefore, KE at the top is: 1/2(1)(20)^2 = 200J. PE at the top is: (1)(10)(20) = 200J

Total ME at the top of the loop is: 400J. Final KE is: 400J = 1/2mv^2; sqrt(800) = v=~28m/s

3.) Consider the forces at the top of the loop:

a. Normal Force points Downwards

b. Gravity points Downwards

Centripetal Force (net force) also points down. Therefore: mv^2/r = N + mg

Plugging in the appropriate variables and solving for mv^2, we get: (80N + 10N)10

900N = mv^2 divide this by 2 to get: 1/2mv^2 (KE)

450J = KE at the top of the loop.

PE at the top of the loop equals: mgh (1x10x20m) = 200J

Therefore, using conservation of ME, the final velocity equals:

PEi + KEi = KEf (note: PEf = 0J since we're at the bottom of the loop)

650J = KEf

650J = 1/2mv^2

1300J / 1kg = v^2

sqrt(1300) = v

~36 m/s