Final answer:
When Rawly descends 21.0 m down the hill, his remaining gravitational potential energy is 6623.3 J and his kinetic energy is 16292.4 J.
Step-by-step explanation:
When Rawly is sledding down the hill, the gravitational potential energy (PEgrav) he has at the top of the hill is being converted into kinetic energy (KE) as he descends. Initially, all of his energy is potential energy, which is given by the equation PEgrav = mgh, where m is Rawly's mass, g is the acceleration due to gravity (9.8 m/s2), and h is the height above the ground.
At the top of the hill (29.5 m tall), his total energy (which is all potential energy in this case) is:
PEgrav,top = mgh
= (79.0 kg)(9.8 m/s2)(29.5 m)
= 22915.7 J
After descending 21.0 m down the hill, the height h is now (29.5 m - 21.0 m) = 8.5 m.
Therefore, his remaining gravitational potential energy is:
PEgrav,current = mgh
= (79.0 kg)(9.8 m/s2)(8.5 m)
= 6623.3 J
Since energy is conserved (ignoring any work done by friction), the loss in gravitational potential energy equals the gain in kinetic energy.
Therefore, the kinetic energy KE he has after descending 21.0 m would be:
KE = PEgrav,top - PEgrav,current
= 22915.7 J - 6623.3 J
= 16292.4 J
At this point, Rawly has 6623.3 J of gravitational potential energy and 16292.4 J of kinetic energy.