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)
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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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 )
[tex]\texttt{ }[/tex]
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 !
[tex]\texttt{ }[/tex]
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
[tex]\texttt{ }[/tex]
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]
[tex]\texttt{ }[/tex]
[tex]\texttt{ }[/tex]
Grade: High School
Subject: Physics
Chapter: Circular Motion
