A car of mass 1100 kg starts from rest at sea level and climbs a hill of altitude 50 m. At the

top of the hill the car has a speed of 25 m/s. From the top of the hill the driver turns off the
engine and coasts down to an altitude of 15 m. Assume the friction and the air resistance to be
negligibly small.
A. What is the speed of the car when the altitude is 15m?
B. ) After passing the altitude of 15m, the driver climbs up again, without turning the engine on. In
this case, the speed of the car would be zero at an altitude of _____?

A. To find the speed of the car when the altitude is 15m, we can use the conservation of energy principle. At the top of the hill, the car has potential energy due to its height and kinetic energy due to its speed. As the car coasts down, it loses potential energy but gains kinetic energy. The total mechanical energy of the car remains constant.

Let's calculate the potential energy at the top of the hill using the formula:

Potential energy = mass * acceleration due to gravity * height

Potential energy = 1100 kg * 9.8 m/s^2 * 50 m
Potential energy = 539,000 J

Using the conservation of energy principle, the total mechanical energy (E) is equal to the sum of potential energy (PE) and kinetic energy (KE):

E = PE + KE

At the top of the hill, all the energy is in the form of potential energy, so E = PE.

At the altitude of 15m, all the potential energy has been converted to kinetic energy:

E = KE

Therefore, we can equate the potential energy at the top of the hill to the kinetic energy at the altitude of 15m:

539,000 J = (1/2) * mass * velocity^2

Rearranging the equation to solve for velocity:

velocity = sqrt((2 * 539,000 J) / mass)
velocity ≈ sqrt((2 * 539,000 J) / 1100 kg)
velocity ≈ sqrt(980 J/kg)
velocity ≈ 31.3 m/s

So, the speed of the car when the altitude is 15m is approximately 31.3 m/s.

B. After passing the altitude of 15m, the driver climbs up again, without turning the engine on. In this case, the speed of the car would be zero at an altitude of 15m.
Since there is no additional external force applied to the car, it will gradually slow down and eventually come to a stop at an altitude of 15m.

In order to find the answer to these questions, we need to apply the conservation of energy principle.

A. To find the speed of the car when the altitude is 15m, we can equate the initial potential energy to the final kinetic energy.

The initial potential energy is given by mgh, where m represents the mass of the car (1100 kg), g represents the acceleration due to gravity (approximately 9.8 m/s^2), and h represents the initial altitude (50 m).

The final kinetic energy is given by (1/2)mv^2, where v represents the final velocity of the car (25 m/s).

Since the friction and air resistance are negligible, we can assume that there is no loss of energy. Therefore, we can set the initial potential energy equal to the final kinetic energy:

mgh = (1/2)mv^2

Simplifying the equation, we have:

1100 kg * 9.8 m/s^2 * 50 m = (1/2) * 1100 kg * v^2

Solving for v, we get:

v = √(2 * 9.8 m/s^2 * 50 m) ≈ 31.3 m/s

Therefore, the speed of the car when the altitude is 15m is approximately 31.3 m/s.

B. To find the altitude at which the speed of the car would be zero, we need to consider the conservation of mechanical energy when the car is climbing back up the hill.

Considering the initial potential energy at an altitude of 15m and the final kinetic energy being zero, we can write:

mgh = (1/2)mv^2

Simplifying the equation and rearranging for h, we have:

h = (1/2) * v^2 / g

Plugging in the values, where v is zero and g is 9.8 m/s^2, we get:

h = (1/2) * 0 / 9.8 ≈ 0 m

Therefore, the speed of the car would be zero at an altitude of approximately 0 meters (at sea level).