A truck of mass 1200kg is parked at the top of a hill, 150m high. The truck driver lets the truck run freely down the hill to the bottom. A) What is the maximum velocity that the truck can achieve at the bottom of the hill?
the gravitational potential becomes kinetic energy
1/2 m v^2 = m g h ... v = √(2 g h)
To determine the maximum velocity that the truck can achieve at the bottom of the hill, we can use the principle of conservation of energy.
The truck starts at rest at the top of the hill, so its initial kinetic energy is zero. Its total energy at the top of the hill is given by the sum of its gravitational potential energy and its initial kinetic energy:
E_initial = PE_top + KE_initial
The potential energy at the top of the hill is given by the product of the truck's mass (m = 1200 kg), the acceleration due to gravity (g = 9.8 m/s^2), and the height of the hill (h = 150 m):
PE_top = m * g * h
Since the truck is at rest at the top of the hill, its initial kinetic energy is zero:
KE_initial = 0
Therefore, the total energy at the top of the hill is:
E_initial = m * g * h
At the bottom of the hill, all of the truck's initial potential energy is converted into kinetic energy:
E_final = KE_final
The kinetic energy at the bottom of the hill is given by the equation:
KE_final = (1/2) * m * v^2
where v is the final velocity of the truck at the bottom of the hill.
Equating the initial and final energies, we have:
E_initial = E_final
m * g * h = (1/2) * m * v^2
Canceling out the mass and simplifying the equation, we get:
g * h = (1/2) * v^2
Now, we can solve for v:
v^2 = 2 * g * h
Taking the square root of both sides, we get:
v = √(2 * g * h)
Substituting the given values:
v = √(2 * 9.8 * 150)
v ≈ 54.33 m/s
Therefore, the maximum velocity that the truck can achieve at the bottom of the hill is approximately 54.33 m/s.