A 100 kg car is traveling at 20m/s over the top of a circular hill as shown below. The radius of the circle is 200 m.

What force does the ground exert on the car at the top of the hill?
After reaching the top of the hill, the car descends a vertical distance of 20 m. What is the final velocity assuming that there is no friction?

suspect 1000 kg not 100 kg but anyway

gravity down = 9.81 *100 = 981 N
acceleration down = v^2/r

Up force = F
so
981 - F = ma = 100 v^2/r
so
F = 981 - 100 (400)/200
= 781 N

but as you know if it is a real car I think 7810 N

To find the force exerted by the ground on the car at the top of the hill, we need to consider the centripetal force and the gravitational force.

The centripetal force is given by the formula:
Fc = m * v^2 / 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 circular path.

Plugging in the values:
Fc = (100 kg) * (20 m/s)^2 / 200 m
Fc = 1000 N

So, the force exerted by the ground on the car at the top of the hill is 1000 N.

To find the final velocity of the car after descending a vertical distance of 20 m, we can use the principle of conservation of mechanical energy.

The initial mechanical energy of the car at the top of the hill is given by:
Ei = m * g * h
where Ei is the initial mechanical energy, m is the mass of the car, g is the acceleration due to gravity, and h is the height.

Plugging in the values:
Ei = (100 kg) * 9.8 m/s^2 * 20 m
Ei = 19600 J

The final mechanical energy of the car at the bottom of the hill is given by:
Ef = m * g * h' + 0.5 * m * vf^2
where Ef is the final mechanical energy, m is the mass of the car, g is the acceleration due to gravity, h' is the vertical distance, and vf is the final velocity.

Plugging in the values:
Ef = (100 kg) * 9.8 m/s^2 * 0 m + 0.5 * (100 kg) * vf^2
Ef = 0 + 50 vf^2

Since there is no friction, the mechanical energy is conserved, so we can set Ei equal to Ef:
19600 J = 50 vf^2

Solving for vf, we get:
vf^2 = 19600 J / 50 kg
vf^2 = 392 m^2/s^2

Taking the square root of both sides, we get:
vf = √392 m/s
vf ≈ 19.8 m/s

So, assuming no friction, the final velocity of the car after descending a vertical distance of 20 m is approximately 19.8 m/s.

To determine the force exerted by the ground on the car at the top of the hill, we need to consider the forces acting on the car at that point. In this case, there are two forces at play: the gravitational force and the normal force.

1. Gravitational force: This force is acting towards the center of the circular path and can be calculated using the formula F_gravity = m * g, where m represents the mass of the car (100 kg) and g is the acceleration due to gravity (approximately 9.8 m/s^2).

F_gravity = (100 kg) * (9.8 m/s^2) = 980 N

2. Normal force: This force acts perpendicular to the surface of the hill and counters the gravitational force to keep the car on the circular path. At the top of the hill, the normal force equals the gravitational force plus a centripetal force.

Centripetal force is given by the formula F_centripetal = (m * v^2) / r, where v is the velocity of the car (20 m/s) and r is the radius of the circular path (200 m).

F_centripetal = (100 kg * (20 m/s)^2) / 200 m = 1000 N

Therefore, the total force exerted by the ground on the car at the top of the hill is the sum of the gravitational force and the centripetal force:

Total force = F_gravity + F_centripetal = 980 N + 1000 N = 1980 N.

To determine the final velocity of the car after descending a vertical distance of 20 m with no friction, we can use the principle of conservation of energy. At the top of the hill, the car has both kinetic energy (due to its velocity) and potential energy (due to its height).

Initial potential energy at the top = m * g * h, where m is the mass of the car, g is the acceleration due to gravity, and h is the height (20 m).

Initial potential energy = (100 kg) * (9.8 m/s^2) * (20 m) = 19600 J.

Since there is no friction, the total mechanical energy (sum of kinetic and potential energy) is conserved throughout the motion.

Therefore, the final mechanical energy equals the initial mechanical energy:

Final mechanical energy = Initial mechanical energy
0.5 * m * v^2 + m * g * h = 19600 J.

Since the car descends vertically, its final potential energy is zero. Thus, the equation becomes:
0.5 * m * v^2 = 19600 J.

Rearranging the equation to solve for v (final velocity):
v = sqrt((2 * 19600 J) / m).

Substituting the given values:
v = sqrt((2 * 19600 J) / 100 kg) = sqrt(392 J/kg) ≈ 19.8 m/s.

Therefore, the final velocity of the car, assuming no friction, after descending a vertical distance of 20 m is approximately 19.8 m/s.