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3 years ago

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Holding onto a tow rope moving parallel to a frictionless ski slope, a 61.8 kg skier is pulled up the slope, which is at an angl

e of 6.8° with the horizontal. What is the magnitude Frope of the force on the skier from the rope when (a) the magnitude v of the skier's velocity is constant at 2.47 m/s and (b) v = 2.47 m/s as v increases at a rate of 0.109 m/s2?

1 answer:

nata0808 [166]3 years ago

8 0

Answer:

a) Frope= 71.7 N

b) Frope=6.7 N

Explanation:

In the figure the skier is simulated as an object, "a box".

a) At constant velocity we can say that the object is in equilibrium, so we apply the Newton's first law:

∑F=0

Frope=w*sen6.8°

Frope=71.71N

Take into account that w is the weight that is calculated as mass per gravitiy constant:

w=m*g

$w=61.8Kg*9.8\frac{m}{s^{2} }$

$w=605.64N$

b) In this case the system has an acceleration of 0.109m/s2. Then, we apply Newton's second law of motion:

F=m*a

F=61.8Kg*0.109m/s2

Frope=6.73N

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Answer:

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(B). The rotational speed of sphere is 400 rad/s.

Explanation:

Given that,

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Using formula of momentum

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Put the value into the formula

$L=\dfrac{1}{2}\times10\times(9.0\times10^{-2})^2\times320$

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(B). Rotation momentum of sphere is same rotational momentum of the flywheel

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Using formula of rotational momentum

$L_{sphere}=L_{flywheel}$

$I\omega_{sphere}=I\omega_{flywheel}$

$\omega_{sphere}=\dfrac{I\omega_{flywheel}}{I_{sphere}}$

$\omega_{sphere}=\dfrac{I\omega_{flywheel}}{\dfrac{2}{5}mr^2}$

Put the value into the formula

$\omega_{sphere}=\dfrac{12.96}{\dfrac{2}{5}\times10\times(9.0\times10^{-2})^2}$

$\omega_{sphere}=400\ rad/s$

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Explanation:

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Explain

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