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LekaFEV [45]
3 years ago
14

In the first figure here, a sled is held on an inclined plane by a cord pulling directly up the plane. The sled is to be on the

verge of moving up the plane. In the second figure, the magnitude F required of the cord's force on the sled is plotted versus the coefficient of static friction μs between sled and plane: F1 = 2.2 N, F2 = 4.5 N, and μ2 = 0.512. At what angle θ (in ˚) is the plane inclined?
Physics
1 answer:
Charra [1.4K]3 years ago
4 0
When it says something like 'on the verge of moving,' it means that the pulling force and static friction force and gravitational force all cancel out! Any more pulling force and it is ready to move! At some point, you want F as a function of <span>μs</span>, to determine the force needed depending on the coefficient of static friction. This function, <span>F(<span>μs</span>)</span>, will rely on the angle θ as well, but we want to consider just one angle θ in every scenario. One value means it is constant. But if we know the F, and we know <span>μs</span>, we can find what the constant angle θ must be! If F is the pulling force, <span>FS</span> is the static friction force, and <span>FG</span> is gravitational force, <span><span><span>Fnet</span>=0</span><span>=F+<span>FS</span>+<span>FG</span></span><span>=F+<span>FN</span><span>μs</span>+mgsinθ</span><span>=F+mgcosθ<span>μs</span>+mgsinθ</span><span>=0</span></span> Then you can find <span>F(<span>μs</span>)</span>, but then there is the issue of solving for the θ<span> to make it true.</span>
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A 4.40-kilogram hoop starts from rest at a height 1.70 m above the base of an inclined plane and rolls down under the influence
Anestetic [448]

Answer:

The linear velocity is  v=4.08m/s

Explanation:

According to the law of conservation of energy

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Here R is the radius of the hoop

         w is the angular velocity which the hoop has at the bottom of the lower part of the inclined plane which is mathematically represented as

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Where v linear speed of the hoop's center of mass just as the hoop leaves the incline and rolls onto a horizontal surface

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=>           mgh = mv^2

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3 years ago
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A crate with a mass of 110 kg glides through a space station with a speed of 4.0 m/s. An astronaut speeds it up by pushing on it
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Answer:

The final speed of the crate after the astronaut push to slow it down is 4.50 m/s

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<u>Given:  </u>

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<u>Solution  </u>

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Ef=Ei+W                                                 (1)

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In this case, the rest energy doesn't change so we can cancel the rest energy in both sides and substitute with the approximate expression of the kinetic energy of the crate at low speeds where K = 1/2 mv^2 and equation (2) will be in the form

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vi' = vf

In this case, we will apply equation (3) for the second process to be in the

1/2mvf^2=1/2mvi'^2+W'                                 (3*)

The force in the second process is F = 170 N and the displacement is 4 m. The force and the displacement are in the opposite direction, hence the work done is negative and will be calculated by  

W'= —F Δr = —(170N)(4 m)= —680J

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1/2mvf'^2=1/2mvi'^2+W'

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7 0
3 years ago
An object that has kinetic energy must what ?
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