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3 years ago

6

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 ra

mp. 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?

1 answer:

Katarina [22]3 years ago

3 0

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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2) The grinding wheel decelerates until rest is reached.

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$\theta - \theta_{o} = \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}$

If $\omega_{o} = 25\,\frac{rad}{s}$ , $t = 2.40\,s$ and $\alpha = 26\,\frac{rad}{s^{2}}$ , then:

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If $\omega_{o} = 25\,\frac{rad}{s}$ , $t = 2.40\,s$ and $\alpha = 26\,\frac{rad}{s^{2}}$ , then:

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$\Delta \theta = 134.88\,rad + 435\,rad$

$\Delta \theta = 569.88\,rad$

The total angle of the grinding wheel is 569.88 radians.

b) Before finding the instant when the grinding wheel stops, it is needed to find the value of angular deceleration, which can be determined from the following kinematic expression:

$\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha \cdot (\theta-\theta_{o})$

The angular acceleration is now cleared:

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