A 1400kg car starts at rest and rolls down a hill from a height of 12m. (a) how much kinetic energy does the car have at the bottom of the hill?(b) how fast is the car going at the bottom of the hill?
Can someone explain how to do it!
They must expect you to neglect friction. The car would have to be in neutral gear, with well lubricated wheel bearings, and properly inflated tires.
The loss of potential energy, M g H, would then equal the gain in kinetic energy, (1/2)MV^2.
V = sqrt(2*g*H)
H = 12 m
g = 9.8 m/s^2
The car's mass does not matter when computing the velocity, but it does matter when computing kinetic energy. \
Sure! To solve this problem, we will use the principles of potential energy and kinetic energy.
(a) To find the potential energy at the top of the hill, we use the formula: potential energy (PE) = mass (m) * gravitational acceleration (g) * height (h).
In this case, the mass of the car is 1400 kg, the height of the hill is 12 m, and the gravitational acceleration is approximately 9.8 m/s^2.
PE = 1400 kg * 9.8 m/s^2 * 12 m
= 164,640 J
Since energy is conserved, all the potential energy at the top of the hill will be converted into kinetic energy at the bottom of the hill.
So, the kinetic energy (KE) at the bottom of the hill will also be 164,640 J.
(b) To find the velocity (speed) of the car at the bottom of the hill, we can use the formula: kinetic energy (KE) = 0.5 * mass (m) * velocity (v)^2.
In this case, the mass of the car is still 1400 kg, and the kinetic energy is 164,640 J.
KE = 0.5 * 1400 kg * v^2
Rearranging the formula, we get:
v^2 = (2 * KE) / m
v^2 = (2 * 164,640 J) / 1400 kg
v^2 = 235.2 m^2/s^2
Taking the square root of both sides gives us:
v = √235.2 m/s
v ≈ 15.3 m/s
So, the car will be going at approximately 15.3 m/s at the bottom of the hill.
To summarize:
(a) The car will have 164,640 J of kinetic energy at the bottom of the hill.
(b) The car will be traveling at approximately 15.3 m/s at the bottom of the hill.