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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 ground 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.)

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Answer:

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

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

Step-by-step 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) ((F_(y) )/(F_(x) ) )


\beta = tan^(-1) ((10.9*10^(-6) )/(0.340*10^(-6) ) )

β = 88.21°

A flea jumps by exerting a force of 1.09 ✕ 10−5 N straight down on the ground. A breeze-example-1
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