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4 years ago

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A flea jumps by exerting a force of 1.09 ✕ 10−5 N straight down on the ground. A breeze blowing on the flea parallel to the grou

nd exerts a force of 0.340 ✕ 10−6 N on the flea. Find the direction and magnitude of the acceleration of the flea if its mass is 6.10 ✕ 10−7 kg. Do not neglect the gravitational force. (Assume that the breeze blows in the +x-direction and that the +y-direction is up.)

1 answer:

patriot [66]4 years ago

5 0

Answer:

a= 17.877 m/s² : Magnitude of the acceleration of the flea

β = 88.21° : Direction of the acceleration of the flea

Explanation:

Conceptual analysis

We apply Newton's second law:

∑F = m*a (Formula 1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Problem development

Look at the flea free body diagram in the attached graphic

The acceleration is presented in the direction of the resultant force (R) applied over the flea .

$R= \sqrt{(F_{x})^{2} + (F_{y})^{2} }$

$R= \sqrt{(10.9)^{2}+(0.340)^{2} } *10^{-6} N$

R= 10.905*10⁻⁶ N

We apply the formula (1) to calculate the magnitude of the acceleration of the flea

∑F = m*a m = 6.1 * 10⁻⁷ kg

R = m*a

a= R/m

a= (10.905*10⁻⁶) / (6.1 * 10⁻⁷ )

a= 17.877 m/s²

β: Direction and magnitude of the acceleration of the flea

$\beta = tan^{-1} (\frac{F_{y} }{F_{x} } )$

$\beta = tan^{-1} (\frac{10.9*10^{-6} }{0.340*10^{-6} } )$

β = 88.21°

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3 years ago

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Answer:

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After doing an analysis of the free-body diagram, we can build our set of equations by using Newton's thrid law:

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we can divide both sides of the equation into mg so we get:

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as you may see, the angle doesn't depend on the mass of the box, only on the static coefficient of friction. When solving for $\mu_{s}$ we get:

$\mu_{s}=\frac{sin \alpha}{cos \alpha}$

when simplifying this we get that:

$\mu_{s}=tan \alpha$

now we can solve for the angle so we get:

$\alpha= tan^{-1}(\mu_{s})$

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$\alpha= tan^{-1}(0.350)$

which yields:

α=19.29°

which rounds to:

α=19°

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