Question

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.)

Answer 1

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°