A 250.0 kg roller coaster car has 20000 J of potential energy at the top of a hill. Neglecting frictional losses, what is the velocity of the car at the bottom of the hill?
The KE will equal the PE at the top
1/2 m v^2= 20000
solve for v.
Could you please put this in layman's terms?
I believe that this answer is 9.8 m/s am I right?
The kinetic energy (1/2 *mass*velocity squared) will equal the potential energy a the top
1/2 m v2 = 20000
The veloicity will be near 13 meters/second at the bottom
To find the velocity of the roller coaster car at the bottom of the hill, we can apply the principle of conservation of energy. According to this principle, the total mechanical energy of an object remains constant in the absence of external forces. In this case, the roller coaster car only has gravitational potential energy at the top of the hill, which will be converted to kinetic energy at the bottom of the hill.
The potential energy (PE) at the top of the hill can be calculated using the formula:
PE = m * g * h
where m is the mass of the car (250.0 kg), g is the acceleration due to gravity (9.8 m/s²), and h is the height of the hill. However, the height of the hill is not given in the question. Therefore, we cannot calculate potential energy directly.
However, we are given that the potential energy at the top of the hill is 20,000 J.
So, we can set up an equation by equating the potential energy at the top of the hill to the kinetic energy at the bottom of the hill, since mechanical energy is conserved:
PE = KE
PE = m * g * h
KE = (1/2) * m * v²
where v is the velocity of the car at the bottom.
Equating the two equations, we get:
m * g * h = (1/2) * m * v²
Simplifying the equation, we can cancel out the mass "m" on both sides:
g * h = (1/2) * v²
Now, we need to rearrange the equation to solve for v:
v² = 2 * g * h
Taking the square root of both sides:
v = √(2 * g * h)
Since we don't have the height of the hill, we cannot calculate the exact velocity.