a 1000kg car goes over the top of a hill with a radius curvature equal to 40.0m. At the top of the hill the normal force is 2,500 N. How fast is the car going?
normal force=mv^2/r-mg
solve for v
To find the speed of the car at the top of the hill, we can use the concept of centripetal force. The centripetal force acting on an object moving in a circle is equal to the product of the mass of the object, its speed squared, and the radius of curvature of its path divided by the distance between the center of the circle and the object. In this case, we can calculate the speed as follows:
1. First, we need to find the net force acting on the car at the top of the hill. Since the car is not accelerating vertically (up and down motion), the sum of the vertical forces must be zero. The gravitational force pulling the car downward will be partially balanced by the normal force acting upward. Therefore, we have:
Net force = Weight - Normal force
The weight of the car can be calculated using the equation: Weight = mass × gravitational acceleration, where the gravitational acceleration is approximately 9.8 m/s².
Weight = 1000 kg × 9.8 m/s² = 9800 N
Net force = 9800 N - 2500 N = 7300 N
2. The net force acting on the car will be the centripetal force at the top of the hill. Therefore, we have:
Centripetal force = Net force = (mass × speed²) / radius
Rearranging the equation, we get:
speed² = (Centripetal force × radius) / mass
3. Plugging in the given values, we have:
speed² = (7300 N × 40.0 m) / 1000 kg
Simplifying the equation:
speed² = 292000 N·m / 1000 kg
speed² = 292 m²/s²
Taking the square root of both sides, we get:
speed = √(292 m²/s²) ≈ 17.1 m/s
Therefore, the car is going approximately 17.1 m/s at the top of the hill.