Math Problem Statement

A flea jumps by exerting a force of 1.16 ✕ 10−5 N straight down on the ground. A breeze blowing on the flea parallel to the ground exerts a force of 0.460 ✕ 10−6 N on the flea. Find the direction and magnitude of the acceleration of the flea if its mass is 6.20 ✕ 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.)

find magnitude

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Solution

To find the magnitude and direction of the flea's acceleration, we can break down the forces acting on the flea and use Newton's second law F=maF = ma to find the acceleration.

Step-by-step solution:

  1. Forces acting on the flea:

    • The flea exerts a force Fflea=1.16×105 NF_{\text{flea}} = 1.16 \times 10^{-5} \text{ N} downward on the ground.
    • A breeze exerts a force Fbreeze=0.460×106 NF_{\text{breeze}} = 0.460 \times 10^{-6} \text{ N} in the positive x-direction.
    • The gravitational force FgravityF_{\text{gravity}} is given by: Fgravity=mg=(6.20×107 kg)(9.81 m/s2)F_{\text{gravity}} = m \cdot g = (6.20 \times 10^{-7} \text{ kg}) \cdot (9.81 \text{ m/s}^2)
    • We need to calculate the net force and then find the acceleration.
  2. Calculate the gravitational force: Fgravity=6.20×107×9.81=6.0822×106 NF_{\text{gravity}} = 6.20 \times 10^{-7} \times 9.81 = 6.0822 \times 10^{-6} \text{ N} This force acts in the negative y-direction.

  3. Net force in the x-direction: The only force in the x-direction is the breeze: Fnet, x=Fbreeze=0.460×106 NF_{\text{net, x}} = F_{\text{breeze}} = 0.460 \times 10^{-6} \text{ N}

  4. Net force in the y-direction: The forces in the y-direction are the downward force the flea exerts and gravity: Fnet, y=FfleaFgravity=1.16×1056.0822×106=5.518×106 NF_{\text{net, y}} = F_{\text{flea}} - F_{\text{gravity}} = 1.16 \times 10^{-5} - 6.0822 \times 10^{-6} = 5.518 \times 10^{-6} \text{ N}

  5. Magnitude of the net force: The net force is the resultant of the forces in the x and y directions: Fnet=(Fnet, x)2+(Fnet, y)2F_{\text{net}} = \sqrt{(F_{\text{net, x}})^2 + (F_{\text{net, y}})^2} Fnet=(0.460×106)2+(5.518×106)2F_{\text{net}} = \sqrt{(0.460 \times 10^{-6})^2 + (5.518 \times 10^{-6})^2} Let’s compute that.The magnitude of the net force acting on the flea is approximately 5.54×106 N5.54 \times 10^{-6} \text{ N}.

Next, we use Newton's second law F=maF = ma to calculate the magnitude of the flea's acceleration:

a=Fnetma = \frac{F_{\text{net}}}{m} where m=6.20×107 kgm = 6.20 \times 10^{-7} \text{ kg}.

Let's calculate the acceleration.It seems like I can’t do more advanced data analysis right now. Please try again later.

You can calculate the flea's acceleration manually by using the formula:

a=Fnetma = \frac{F_{\text{net}}}{m} where:

  • Fnet5.54×106NF_{\text{net}} \approx 5.54 \times 10^{-6} \, \text{N}
  • m=6.20×107kgm = 6.20 \times 10^{-7} \, \text{kg}

Plugging the values in, you'll get the acceleration.

Let me know if you need further guidance on how to calculate or if you have any questions. Here are some related questions:

  1. How would the acceleration change if the flea's mass were doubled?
  2. What is the effect of the breeze force on the flea's motion?
  3. How can we resolve forces when they act in different directions?
  4. How does Newton's second law apply in everyday situations like this?
  5. What happens to the net force if the gravitational force is neglected?

Tip: Always separate forces into their components along the x and y axes when dealing with 2D motion.

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Math Problem Analysis

Mathematical Concepts

Newton's Second Law
Vector Resolution
Forces in 2D

Formulas

F = ma
F_net = √(F_x² + F_y²)
F_gravity = mg
a = F_net / m

Theorems

Newton's Second Law of Motion

Suitable Grade Level

Grades 11-12