A 12.0 kg sheep, starting at rest, is accelerated upward by a helicopter. The ca

Question

A 12.0 kg sheep, starting at rest, is accelerated upward by a helicopter. The cable has a constant tension of 180 N. The sheep climbs 5.0 m. Compute The work done on the sheep by the cable in raising the sheep 5.0 m. The increase in potential energy of the sheep in rising 5.0 m. The final kinetic energy of the sheep in being lifted to a height of 5.0 m. No let's do the problem the old way. Compute The acceleration of the sheep. The speed of the sheep after climbing 5m. The final kinetic energy. Does it match your answer to (c)? Answers to starred questions

Explanation / Answer

(a)W = Fy

Force is 180N from the constant tension in the rope and it rises 5m. The sheep moves in the same direction as the rope so it's positive work

W = 180N (5m) = 900J

(b)In small distances near the surface of the earth,

U = mgh where h is the difference in height

U = (12kg)(9.8m/s2)(5m) = 588J

(c) K = Wnet

The helicopter does positive work on the sheep, however, gravity does negative work on the sheep as well.

Fg = mg

so Wnet is Wh + Wg which is FTy + Fgy = FTy + (-mgy) = y(FT - mg)

Wnet = 5m (180N - 12kg(9.8m/s2))

Wnet = 312J = K

K = 312J

(d) F = ma

a = F/m

The force is the total net force, so the Tension of the string minus the force of gravity

a = (T - mg)/m = (180N - 9.8m/s2(12kg)) / 12kg

a = 5.2 m/s2

(e) There is constant acceleration so use one of the constant acceleration equations

v2 = v02 + 2a(y - y0)

v2 = 0 + 2a(y - 0)

v2 = 2ay

v = 2ay

v = (2(5.2m/s2)(5m)) = 7.2m/s2

(f) K = 0.5mv2

K = 0.5 (12kg) (7.2m/s2)2

K 312J

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