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 _____?
mass is irrelevant - cancels everywhere
Energy at top = m g h + (1/2) m v^2
= (9.8 * 50 + 625/2)m
= 803 m Joules
Energy at 15 = m ( 9.8*15 + v^2/2) = 803 m
v^2/2 = 656
v = 36.2 m/s
coasting up fro 15, still same total energy
m g h =803 m
9.8 h = 803
h = 81.9 meters
Thanks!!
15m/s to 81.9meters
To solve this problem, we can use the principles of conservation of energy.
A. To find the speed of the car when the altitude is 15m, we need to first calculate the potential energy at the top of the hill and convert it to kinetic energy at a lower altitude.
The potential energy (PE) of an object at a certain altitude is given by the equation: PE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the altitude.
Given that the mass of the car is 1100 kg, the altitude is 50m, and neglecting air resistance and friction, we can calculate the initial potential energy at the top of the hill:
PE1 = mgh = (1100 kg)(9.8 m/s^2)(50 m) = 539,000 J
Now, using the principle of conservation of energy, we know that the total mechanical energy of the system (car) is conserved. Therefore, the sum of the potential energy and the kinetic energy at the top of the hill is equal to the sum of the potential energy and the kinetic energy at the lower altitude.
At the top of the hill, the car has a speed (v1) of 25 m/s and zero potential energy. Therefore, the total mechanical energy at the top of the hill is:
ME1 = KE1 + PE1 = 0 + 539,000 J = 539,000 J
At the altitude of 15m, the potential energy (PE2) is given by:
PE2 = mgh = (1100 kg)(9.8 m/s^2)(15 m) = 161,700 J
Since air resistance and friction are negligible, the total mechanical energy at this lower altitude is:
ME2 = KE2 + PE2
Therefore, the kinetic energy (KE2) at the lower altitude can be calculated as:
KE2 = ME2 - PE2 = 539,000 J - 161,700 J = 377,300 J
Using the equation for kinetic energy, KE = 1/2 mv^2, we can solve for the speed (v2) at the altitude of 15m:
1/2 mv2^2 = 377,300 J
Simplifying the equation:
v2^2 = (2 * 377,300 J) / (1100 kg) = 686 m^2/s^2
Taking the square root of both sides to solve for v2:
v2 ≈ 26.2 m/s
Therefore, the speed of the car when the altitude is 15m is approximately 26.2 m/s.
B. After passing the altitude of 15m, the car will start climbing up again. Since the driver does not turn on the engine, the initial kinetic energy will gradually be converted to potential energy as the car goes higher. At any altitude, the sum of the kinetic energy and potential energy will be equal to the total mechanical energy, which is conserved.
When the car reaches an altitude where the speed is zero, it means that all of the initial kinetic energy has been converted back to potential energy.
So, at an altitude where the speed of the car would be zero, the potential energy would be equal to the initial mechanical energy at the altitude of 15m:
PE3 = ME2 = 377,300 J
Using the equation for potential energy, PE = mgh, we can solve for the altitude (h3):
mgh3 = 377,300 J
h3 = 377,300 J / (1100 kg * 9.8 m/s^2)
h3 ≈ 34.4 m
Therefore, the speed of the car would be zero at an altitude of approximately 34.4 m.