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kolbaska11 [484]

4 years ago

12

How much work did the movers do (horizontally) pushing a 43.0-kg crate 10.4 m across a rough floor without acceleration, if the

effective coefficient of friction was 0.60

1 answer:

Ilia_Sergeevich [38]4 years ago

8 0

The crate is in equilibrium. Newton's second law gives

∑ <em>F</em> (vertical) = <em>n</em> - <em>mg</em> = 0

∑ <em>F</em> (horizontal) = <em>p</em> - <em>f</em> = 0

where

• <em>n</em> = magnitude of the normal force

• <em>mg</em> = weight of the crate

• <em>p</em> = mag. of push exerted by movers

• <em>f</em> = mag. of kinetic friciton, with <em>f</em> = 0.60<em>n</em>

<em />

It follows that

<em>p</em> = <em>f</em> = 0.60<em>mg</em> = 0.60 (43.0 kg) <em>g</em> = 252.84 N

so that the movers perform

<em>W</em> = <em>p</em> (10.4 m) ≈ 2600 J

of work on the crate. (The <em>total</em> work done on the crate, on the other hand, is zero because the net force on the crate is zero.)

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We notice that block-spring system depicts a Simple Harmonic Motion, whose equation of motion is:

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If we know that $m=1.65\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ and $\Delta y = 0.260\,m$ , then the spring constant is:

$k = \frac{(1.65\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{0.260\,m}$

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If we know that $A = 0.130\,m$ , $k = 62.237\,\frac{N}{m}$ , $m=1.65\,kg$ , $x(t) = 0\,m$ and $\phi = 0\,rad$ , then (1) is reduced into this form:

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$\cos (6.142\cdot t) = 0$

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Answer:

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