Answer :
Final answer:
To stop the mass, a force of 595 N is required. The coefficient of kinetic friction of the track is approximately 0.027.
Explanation:
To find the magnitude of the force required to stop the mass, we can use Newton's second law, which states that force is equal to mass multiplied by acceleration. In this case, the acceleration is equal to the rate at which the speed decreases. Therefore, we can use the formula F = ma to calculate the force. Given that the mass is 170 kg and the acceleration is -3.5 m/s² (negative because the force is opposing motion), we can substitute these values into the formula to find the force required:
F = m * a
F = 170 kg * (-3.5 m/s²)
F = -595 N
Therefore, the magnitude of the force required is 595 N.
To find the coefficient of kinetic friction, we can use the formula f= MkN, where f is the force of kinetic friction and N is the normal force. The normal force can be calculated by finding the component of the weight perpendicular to the slope. Since the slope is horizontal, the normal force is equal to the weight of the mass. Substituting the given values into the formula:
f = MkN
45 N = Mk * 170 kg * 9.8 m/s²
Solving for Mk:
Mk = 45 N / (170 kg * 9.8 m/s²)
Mk ≈ 0.027
Therefore, the coefficient of kinetic friction of the track is approximately 0.027.
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Final Answer;
1. The magnitude of the force required to bring the huge and its rider to a stop at a constant rate of 3.5 m/s² is 595 N.
2. The coefficient of kinetic friction of the track is approximately 0.35.
Explanation:
To find the magnitude of the force required to stop the huge and its rider, we can use Newton's second law: \(F = ma\), where \(F\) is the force, \(m\) is the total mass, and \(a\) is the acceleration. In this case, the acceleration is given as 3.5 m/s², and the total mass is 170 kg. So,
[tex]1. \(F = (170 kg)(3.5 m/s²) = 595 N\).[/tex]
To determine the coefficient of kinetic friction (\(\mu_k\)) of the track, we can use the following equation that relates the frictional force (\(f_k\)) to the normal force (\(N\)) and the coefficient of kinetic friction:
[tex]2. \(f_k = \mu_k \cdot N\)[/tex]
The normal force is the force exerted by the track in the upward direction to support the weight of the object. We can find it using the formula:
[tex]3. \(N = mg\)[/tex]
Where \(m\) is the total mass (170 kg) and \(g\) is the acceleration due to gravity (approximately 9.81 m/s²).
[tex]4. \(N = (170 kg)(9.81 m/s²) \approx 1667.7 N\)[/tex]
Now, we can use equation (2) to find \(\mu_k\):
[tex]5. \(f_k = \mu_k \cdot N\)6. \(595 N = \mu_k \cdot 1667.7 N\)7. \(\mu_k = \frac{595 N}{1667.7 N} \approx 0.35\)[/tex]
So, the coefficient of kinetic friction of the track is approximately 0.35.
Understanding these calculations is important in physics to determine the forces and coefficients involved in various scenarios.
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