Question

A car with a mass of \(1000 .\) kg goes over a hill at a constant speed of \(60.0 \mathrm{~m} / \mathrm{s}\). The top of the hill can be approximated as an arc length of a circle with a radius of curvature of \(370 . \mathrm{m}\). What force does the car exert on the hill as it passes over the top?

Short answer

Answer: The force exerted by the car on the hill as it passes over the top is 70 N.

Step-by-step solution

Identify the forces acting on the car at the top of the hill

At the top of the hill, there are two forces acting on the car: the force of gravity (weight) acting downwards and the normal force (the force the hill exerts on the car) acting radially towards the center of the circle. Weight of the car: \(W = mg\) where \(m\) is the mass of the car and \(g\) is the acceleration due to gravity (approximated as 9.8 \(\mathrm{m/s^2}\)). Normal force: The normal force acts perpendicular to the surface of the hill and towards the center of the circle (upwards and towards the center of the circle).

Apply Newton's second law of motion

Since the car moves in a circular motion at the top of the hill, we can apply Newton's second law for centripetal acceleration: \(F_{net} = m \cdot a_c\) The net force \(F_{net}\) is equal to the difference between the weight and the normal force acting on the car: \(F_{net} = W - F_{N}\) The centripetal acceleration \(a_c\) can be calculated using the formula: \(a_c = \frac{v^2}{r}\) where \(v\) is the constant speed and \(r\) is the radius of curvature.

Calculate the weight of the car

Calculate the weight of the car using the mass \(m = 1000\) kg and gravitational acceleration \(g = 9.8 \mathrm{m/s^2}\): \( W = mg = 1000 \mathrm{kg} \times 9.8 \mathrm{m/s^2} = 9800 \mathrm{N} \)

Calculate the centripetal acceleration

Calculate the centripetal acceleration using the constant speed \(v = 60.0 \mathrm{m/s}\) and the radius of curvature \(r = 370\) m: \(a_c = \frac{v^2}{r} = \frac{(60.0 \mathrm{m/s})^2}{370 \mathrm{m}} = 9.73 \mathrm{m/s^2}\)

Apply Newton's second law to find the normal force

Use the net force equation, \(F_{net} = W - F_{N}\), and substitute the values for \(F_{net}\) and \(W\): \(m \cdot a_c = W - F_{N}\) Plug in the values for \(m\), \(a_c\), and \(W\): \( (1000 \mathrm{kg}) (9.73 \mathrm{m/s^2}) = 9800 \mathrm{N} - F_{N} \) Now solve for the normal force \(F_{N}\): \(9730 \mathrm{N} = 9800 \mathrm{N} - F_{N}\) \(F_{N} = 9800 \mathrm{N} - 9730 \mathrm{N}\) \(F_{N} = 70 \mathrm{N}\)

Conclusion

The force exerted by the car on the hill as it passes over the top is equal to the normal force \(F_{N} = 70 \mathrm{N}\).