a 2000 kg car rounds a circular turn of radius 20.0 meters. if the road is flat and the coefficient of static friction between the tires and the road is .70, how fast can the car go without skidding
μ = 0.7
N = mg
g = 9.8 m/s^2
m = 2000 kg
a = angular acceleration = v^2/r
F = μN = ma =mv^2/r
v = sqrt(μNr/m) = sqrt(μgr) = 12 m/s
1)first find the FN
Fn = m x g
2)then find the Ff using the Ff = us x Fn
3) use F = ma formula and isolate a
a = F/m
4) finally use a = V^2/r formula
you will then get 11.719...N
hope this helped
Well, let me calculate that for you... Give me a second.
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Okay, I'm back! So, to figure out the maximum speed the car can go without skidding, we need to use the centripetal force equation: Fc = mv²/r, where Fc is the centripetal force, m is the mass of the car, v is the velocity, and r is the radius of the turn.
Since the car is not skidding, the centripetal force is provided by the static friction, so we have Fs = μs * m * g, where Fs is the static friction force and μs is the coefficient of static friction (given as 0.70 in this case), and g is the acceleration due to gravity.
Since the static friction force Fs is equal to the centripetal force Fc, we can set the equations equal to each other and solve for v:
μs * m * g = mv²/r
We can cancel out the mass m from both sides of the equation:
μs * g = v²/r
Now we just need to rearrange the equation to find v:
v = √(μs * g * r)
Plugging in the numbers, we get:
v = √(0.70 * 9.8 m/s² * 20.0 m)
After evaluating the expression, the maximum speed the car can go without skidding is approximately 19.8 m/s.
But hey, remember to always drive responsibly and not rely on clowns for physics advice!
To determine the maximum speed at which the car can go without skidding, we need to consider the centripetal force required to keep the car moving in a circular path and compare it to the maximum static friction force that the tires can provide.
The centripetal force is given by the equation:
F_c = (m * v^2) / r
Where:
F_c is the centripetal force
m is the mass of the car (2000 kg)
v is the velocity of the car
r is the radius of the circular turn (20.0 meters)
The maximum static friction force is given by:
F_friction = μ * N
Where:
F_friction is the maximum static friction force
μ is the coefficient of static friction (0.70)
N is the normal force applied by the road on the car, which is equal to the weight of the car (m * g), where g is the acceleration due to gravity (9.8 m/s^2).
To avoid skidding, the centripetal force must be less than or equal to the maximum static friction force:
F_c ≤ F_friction
(m * v^2) / r ≤ μ * N
Simplifying the equation:
v^2 ≤ (μ * r * g)
Taking the square root of both sides:
v ≤ sqrt(μ * r * g)
Now we can substitute the given values and calculate the maximum speed:
v ≤ sqrt(0.70 * 20.0 * 9.8) ≈ 10.41 m/s
Therefore, the car can go up to approximately 10.41 m/s without skidding.
To find the maximum speed at which the car can go without skidding, we can use the concept of centripetal force. The centripetal force required for circular motion is provided by the friction force between the tires and the road. Therefore, the maximum speed depends on the friction force, which is determined by the coefficient of static friction.
The formula for centripetal force is:
F = m * a
Where:
F is the centripetal force,
m is the mass of the car, and
a is the centripetal acceleration.
The centripetal acceleration can be calculated using the formula:
a = v^2 / r
Where:
v is the velocity of the car, and
r is the radius of the circular turn.
The maximum static friction force can be calculated using the formula:
F_friction = μ * m * g
Where:
μ is the coefficient of static friction,
m is the mass of the car, and
g is the acceleration due to gravity (approximately 9.8 m/s^2).
To find the maximum speed, we need to equate the centripetal force and the maximum static friction force:
F = F_friction
Therefore:
m * a = μ * m * g
Canceling out the mass from both sides:
a = μ * g
From the equation a = v^2 / r, we can solve for v:
v^2 = a * r
v = √(a * r)
Substituting the value of a from the previous equation:
v = √(μ * g * r)
Now we can calculate the maximum speed without skidding using the given values:
μ = 0.70 (coefficient of static friction)
g = 9.8 m/s^2 (acceleration due to gravity)
r = 20.0 meters (radius of the circular turn)
v = √(0.70 * 9.8 * 20.0)
v = √(137.2)
v ≈ 11.7 m/s
Therefore, the maximum speed at which the car can go without skidding is approximately 11.7 meters per second.