A 61 kg skier leaves the end of a ski-jump ramp with a velocity of 27 m/s directed 21° above the horizontal. Suppose that as a result of air drag the skier returns to the ground with a speed of 16 m/s, landing 12 m vertically below the end of the ramp. From the launch to the return to the ground, by how much is the mechanical energy of the skier-Earth system reduced because of air drag?

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

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.

b

step-by-step explanation: a third of 24 is 8, 8 times 3 is 24 so 8 - 2 would be 6