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16 days ago

The angle θ is slowly increased. Write an expression for the angle at which the block begins to move in terms of μs.

Physics

1 answer:

kicyunya [2.2K]16 days ago

4 0

Answer:

$tan \theta = \mu_s$

Explanation:

Aby obiekt był w spoczynku na nachyleniu, wynikowa siła działająca na niego musi wynosić zero. Równanie sił działających w kierunku równoległym do nachylenia jest następujące:

$mg sin \theta - \mu_s R =0$ (1)

gdzie

$mg sin \theta$ to składowa ciężaru równoległa do nachylenia, przy czym m oznacza masę obiektu, g oznacza przyspieszenie grawitacyjne, a $\theta$ to kąt nachylenia

$\mu_s R$ to siła tarcia, z $\mu_s$ jako współczynnikiem tarcia oraz R jako reakcją normalną nachylenia

Równanie sił w kierunku prostopadłym do nachylenia to

$R-mg cos \theta = 0$

gdzie

R to reakcja normalna

$mg cos \theta$ to składowa ciężaru prostopadła do nachylenia

Obliczając R,

$R=mg cos \theta$

I podstawiając do (1)

$mg sin \theta - \mu_s mg cos \theta = 0$

Rearanżując równanie,

$sin \theta = \mu_s cos \theta\\\rightarrow tan \theta = \mu_s$

To jest warunek, przy którym równowaga jest zachowana: gdy tangens kąta staje się większy niż wartość $\mu_s$ , siła tarcia nie jest w stanie zrównoważyć składowej ciężaru równoległej do nachylenia, dlatego obiekt zaczyna zsuwać się w dół.

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Explanation:

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Answer:

The system's energy amounts to 15 J.

Explanation:

Given that,

Energy E = 2.5 J

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The spring constant needs to be computed.

Using the formula for the mechanical energy of the system,

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Substituting the values into the formula:

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Using the mechanical energy formula,

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Substituting into the formula:

$E=\dfrac{1}{2}\times500\times(6\times10^{-2})$

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