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A cyclist competes in a one-lap race around a flat, circular course of radius 140 m . Starting from rest and speeding up at a constant rate throughout the race, the cyclist covers the entire course in 60 s . The mass of the bicycle (including the rider) is 76 kg . What is the magnitude of the net force Fnet acting on the bicycle as it crosses the finish line?

Respuesta :

Khoso123

Answer:

  Fnet = √(37² + 467²) N = 468.5 N(470 N with the two Significant Figures)

Explanation:

We've got three components in this problem:  

1) linear acceleration * mass, in the direction of motion  

2) friction force, directed toward the center of the circular course

3) the normal force, pushing up from the ground

 s = ½at²  

140m * 2 * π = ½ * a * (60s)²

a = 0.49 m/s², so the force in the direction of motion is

F = 76kg * 0.49 m/s² = 37 N

2) At the finish line, his tangential velocity is  

v = at = 0.49m/s² * 60s = 29 m/s, so the friction force (resisting the centripetal force) is  

Ff = mv² / r = 76g * (29m/s)² / 140m = 467 N

3) the normal force, pushing up from the ground

  Fnet = √(37² + 467²) N = 468.5 N(470 N with the two Significant Figures)

johanrusli

The magnitude of the net force F_net acting on the bicycle as it crosses the finish line is about 467 N

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Further explanation

Centripetal Acceleration can be formulated as follows:

[tex]\large {\boxed {a = \frac{ v^2 } { R } }[/tex]

a = Centripetal Acceleration ( m/s² )

v = Tangential Speed of Particle ( m/s )

R = Radius of Circular Motion ( m )

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Centripetal Force can be formulated as follows:

[tex]\large {\boxed {F = m \frac{ v^2 } { R } }[/tex]

F = Centripetal Force ( m/s² )

m = mass of Particle ( kg )

v = Tangential Speed of Particle ( m/s )

R = Radius of Circular Motion ( m )

Let us now tackle the problem !

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Given:

radius of the circular course = R = 140 m

elapsed time = t = 60 s

mass of the bicycle = m = 76 kg

Asked:

net force = F = ?

Solution:

Firstly , we will calculate the final speed of the bicycle as it crosses the finish line:

[tex]d = ( v + u ) \frac{t}{2}[/tex]

[tex]2 \pi R = ( v + u ) \frac{t}{2}[/tex]

[tex]2 \pi R = ( v + 0 ) \frac{t}{2}[/tex]

[tex]2 \pi R = \frac{vt}{2}[/tex]

[tex]\boxed{v = 4 \pi R \div t}[/tex] → Equation A

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Next, we could calculate the magnitude of the net force as follows:

[tex]F = m \frac{v^2}{R}[/tex]

[tex]F = m \frac{(4 \pi R \div t)^2}{R}[/tex] ← Equation A

[tex]F = m \frac{(4 \pi)^2 R}{t^2}[/tex]

[tex]F = 76 \times \frac{(4 \pi)^2 \times 140}{60^2}[/tex]

[tex]\boxed{F \approx 467 \texttt{ N}}[/tex]

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Learn more

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  • The Acceleration Due To Gravity : https://brainly.com/question/4189441

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Answer details

Grade: High School

Subject: Physics

Chapter: Circular Motion

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