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3 years ago

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When a driver hits the brakes, his car decelerates from 50m/s at a uniform rate of 2.0m/s^2. His car stops after covering some d

istance. How long does it take him to stop?

1 answer:

Assoli18 [71]3 years ago

7 0

Please find attached photograph for your answer. Please do comment whether it is useful or not.

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Accelerating charges radiate electromagnetic waves. calculate the wavelength of radiation produced by a proton in a cyclotron wi

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3 years ago

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4 years ago

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Two workers are sliding 300 kg crate across the floor. One worker pushes forward on the crate with a force of 400 N while the ot

almond37 [142]

Answer:

The kinetic coefficient of friction of the crate is 0.235.

Explanation:

As a first step, we need to construct a free body diagram for the crate, which is included below as attachment. Let supposed that forces exerted on the crate by both workers are in the positive direction. According to the Newton's First Law, a body is unable to change its state of motion when it is at rest or moves uniformly (at constant velocity). In consequence, magnitud of friction force must be equal to the sum of the two external forces. The equations of equilibrium of the crate are:

$\Sigma F_{x} = P+T-\mu_{k}\cdot N = 0$ (Ec. 1)

$\Sigma F_{y} = N - W = 0$ (Ec. 2)

Where:

$P$ - Pushing force, measured in newtons.

$T$ - Tension, measured in newtons.

$\mu_{k}$ - Coefficient of kinetic friction, dimensionless.

$N$ - Normal force, measured in newtons.

$W$ - Weight of the crate, measured in newtons.

The system of equations is now reduced by algebraic means:

$P+T -\mu_{k}\cdot W = 0$

And we finally clear the coefficient of kinetic friction and apply the definition of weight:

$\mu_{k} =\frac{P+T}{m\cdot g}$

If we know that $P = 400\,N$ , $T = 290\,N$ , $m = 300\,kg$ and $g = 9.807\,\frac{m}{s^{2}}$ , then:

$\mu_{k} = \frac{400\,N+290\,N}{(300\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}$

$\mu_{k} = 0.235$

The kinetic coefficient of friction of the crate is 0.235.

5 0

3 years ago