Answer:
22kj
Explanation:
set h = 0 at the end of slide.
final height is -12m
initial condition will be Ui = 0
Ki = 1/2mv² = 1/2 x 61 x (27)² = 22234.5J
Final condition is Ui = mgh = 61 x 9.8 x -12 = -7173J
Ki = 1/2mv²
Ki= 1/2 x 61 x (16)² = 7808J
conservation energy says that
Ui + Ki = Uf +Kf +ΔEth
so ΔEth = Ui + Ki - Uf - Kf
ΔEth = 22234.5 - 7808 + 7173
ΔEth = 21600J
ΔEth =22Kj
Answer:
The mechanical energy is reduced by 21600J due to air drag.
Explanation:
This problem involves the principle of the conservation of energy. Mathematically it can be stated as follows.
K1 + U1 +Wother = K2 + U2
Worher = energy in forms other than gravity. = K2–K1 + U2–U1
K1, K2 = Kinetic energies at the start and finish points
U1 and U2 = Potential energies at the start and finish points.
K1 = 1/2mv1², K2 = 1/2mv2², U1 = mgh1, U2 = mgh2
v1 = 27m/s, v2 = 16m/s, h1 = 12m, h2 = 0m,
m = 61kg
Wother = K2 - K1 + U2 - U1
Wother = 1/2mv2²- 1/2mv1² + mgh2 - mgh1
= 1/2m(v2² - v1² ) + mg(h2 –h1)
= 1/2×61(16² – 27²) + 61×9.8(0 – 12)
= –21600J.
The solution to this problem didn't have to necessarily take into consideration the fact that the skier first attempts a projectile motion. The most important guide to take into consideration when solving problems involving energy changes are the start and end points or positions. This help to treat the problem nicely.
