The coefficient of static friction between the tires of a car and a horizontal road is 0.515. As the car brakes, what is the magnitude of the maximum acceleration of the car? Assume the net force on the car is the maximum static friction force.
The accelerating force M*a must not exceed the maxmum static friction force mus*M*g. (mus is the static friction coefficient)
Therefore
a < mus*g
Solve for the maximum acceleration a.
This assumes that the road surface is not banked, as it is on most sharp turns. When a curve is "banked", gravity helps keep the car on the road.
what is the net force in a rock that weighs 500N if the water it displaces is 300N
net force= weight-bouyancy= 500-300 N
To find the maximum acceleration of the car, we need to use the maximum static friction force. The maximum static friction force is given by the formula:
\( f_{\text{static max}} = \mu_s \cdot N \)
where \( \mu_s \) is the coefficient of static friction and \( N \) is the normal force.
The normal force is the force exerted by the road on the car in the upward direction. In this case, it is equal to the car's weight, which can be calculated using the formula:
\( N = m \cdot g \)
where \( m \) is the mass of the car and \( g \) is the acceleration due to gravity (approximately 9.8 m/s²).
Since we are given the coefficient of static friction (\( \mu_s = 0.515 \)) and we know the acceleration due to gravity (\( g = 9.8 \, \text{m/s}^2 \)), we can substitute these values into the equations to find the maximum static friction force:
\( f_{\text{static max}} = 0.515 \cdot (m \cdot g) \)
Now, we know that the net force acting on the car is equal to the product of its mass and acceleration (F = m * a). In this case, the net force is the maximum static friction force (\( f_{\text{static max}} \)). So we can set up the following equation:
\( f_{\text{static max}} = m \cdot a \)
Substituting the expression for \( f_{\text{static max}} \) into the equation:
\( 0.515 \cdot (m \cdot g) = m \cdot a \)
Now, we can solve for the acceleration (\( a \)) by dividing both sides of the equation by the mass (\( m \)):
\( a = 0.515 \cdot g \)
Plugging in the value of acceleration due to gravity (\( g \)):
\( a = 0.515 \cdot 9.8 \, \text{m/s}^2 \)
Calculating:
\( a \approx 5.057 \, \text{m/s}^2 \)
So, the magnitude of the maximum acceleration of the car is approximately 5.057 m/s².