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Contestants on a ski jump push off hard and typically have a speed of 2 m/s as they reach the starting gate at the top of the ramp. For safety, the skiers should have a maximum speed of 30 m/s when they reach the bottom of the ramp. Friction and air resistance will do total work of magnitude 4000 J on a 85 kg skier during a run down the ramp. What is the maximum height of the ramp for which the maximum safe speed will not be exceeded?

Respuesta :

valeriagonzalez0213

Answer:

h = 50.6 m

Explanation:

Principle of work and energy

ΔE = W

where:

ΔE: mechanical energy change (J)

W : work of the non-conservative forces (J)

ΔE = Ef-Ei

Ef : final mechanical energy

Ei : initial mechanical energy

K =(1/2 )mv² :  Kinetic energy (J)

U= mgh   :Potential energy (J)

m: mass (kg)

v : speed (m/s)

h: high (m)

Data

m = 85 kg : mass of the skier

vi = 2 m/s : Initial speed of the skier

vf = 30 m/s : Final speed of the skier

hf = 0 Final high of the skier

W = - 4000 J ;   Work of the friction force and force of the air resistance

g = 9.8 m/s² : acceleration due to gravity

Principle of work and energy to the skier

ΔE = Ef-Ei

Ef = Kf + Uf  = (1/2 )m(vf)² + mg(hf)= (1/2 )85(30)² + mg(0)= (1/2 )85(30)² J

Ei = Ki + Ui = (1/2 )m(vi)² + mg(hi) = ( (1/2 )85(2)² + 85*9.8(h) ) J

ΔE = Wf

Ef - Ei   = Wf

(1/2 )(85)(30)² -( (1/2 )(85)(2)² +( 85)*9.8(h) )  = -4000

We factor 85 and pass it to the other side of the equation to divide

(1/2 )(30)² -  (1/2 )(2)² - 9.8 (h) = -4000/85

450 - 1 - 9.8 (h)  = -47.0588

449 - 9.8 (h)  = -47.0588

449 +47.0588 = 9.8 (h)

496.0588 =9.8 (h)

h =496.0588/ 9.8

h = 50.6 m

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