the coefficient of static friction between the coffee cup and the roof of a car is 0.28, If the maximum acceleration the car can have without causing a cup of coffee (on the top of the car) to slide is 2.7 m/s^.
What is the smallest amount of time in which the person can accelerate the car from rest to 19 and still keep the coffee cup on the roof?
Duplicate question-- already answered.
To find the smallest amount of time in which the person can accelerate the car from rest to 19 m/s while keeping the coffee cup on the roof, we need to consider the forces acting on the cup.
The force of static friction between the cup and the roof of the car is what prevents the cup from sliding. The maximum force of static friction can be calculated using the equation:
\( F_{\text{{friction}}} = \mu_s \cdot F_{\text{{normal}}} \)
where \( \mu_s \) is the coefficient of static friction and \( F_{\text{{normal}}} \) is the normal force acting on the cup.
In this case, the only force acting vertically on the cup is its weight, which can be calculated using:
\( F_{\text{{normal}}} = m \cdot g \)
where \( m \) is the mass of the cup and \( g \) is the acceleration due to gravity (approximately 9.8 m/s^2).
Now, when the car is accelerating, there are two forces acting on the cup: the force of static friction and the force of acceleration. The force of acceleration can be calculated using Newton's second law:
\( F_{\text{{acceleration}}} = m \cdot a \)
where \( m \) is the mass of the cup (assumed to stay constant) and \( a \) is the acceleration of the car.
For the cup to remain on the roof, the force of static friction must be equal to or greater than the force of acceleration. Therefore, we can set up the following inequality:
\( F_{\text{{friction}}} \geq F_{\text{{acceleration}}} \)
Substituting the equations for \( F_{\text{{friction}}} \) and \( F_{\text{{acceleration}}} \), we get:
\( \mu_s \cdot F_{\text{{normal}}} \geq m \cdot a \)
Replacing \( F_{\text{{normal}}} \) with \( m \cdot g \), we have:
\( \mu_s \cdot m \cdot g \geq m \cdot a \)
We can cancel out the mass of the cup on both sides of the inequality, resulting in:
\( \mu_s \cdot g \geq a \)
Now we have the maximum acceleration that the car can have without causing the cup to slide, which is given as 2.7 m/s^2. Therefore, we have:
\( \mu_s \cdot g \geq 2.7 \, \text{{m/s}}^2 \)
Now we can solve for the coefficient of static friction:
\( \mu_s \geq \frac{{2.7 \, \text{{m/s}}^2}}{{g}} \)
\( \mu_s \geq \frac{{2.7 \, \text{{m/s}}^2}}{{9.8 \, \text{{m/s}}^2}} \)
\( \mu_s \geq 0.27 \)
Therefore, the coefficient of static friction (\( \mu_s \)) must be equal to or greater than 0.27 for the cup to stay on the roof.
Note: We assumed that \( \mu_s \) remains constant throughout the entire acceleration process.
To calculate the smallest amount of time in which the car can accelerate from rest to 19 m/s while keeping the cup on the roof, we need to find the time it takes for the car to reach 19 m/s with an acceleration of 2.7 m/s^2.
We can use the following equation of motion:
\( v = u + at \)
where:
- \( v \) is the final velocity (19 m/s in this case)
- \( u \) is the initial velocity (0 m/s as the car starts from rest)
- \( a \) is the acceleration (2.7 m/s^2 as given)
- \( t \) is the time taken
Rearranging the equation, we get:
\( t = \frac{{v - u}}{{a}} \)
Substituting the values into the equation, we have:
\( t = \frac{{19 \, \text{{m/s}} - 0 \, \text{{m/s}}}}{{2.7 \, \text{{m/s}}^2}} \)
\( t = \frac{{19}}{{2.7}} \) seconds
\( t \approx 7.04 \) seconds
Therefore, the smallest amount of time in which the person can accelerate the car from rest to 19 m/s while keeping the coffee cup on the roof is approximately 7.04 seconds.