a toy car rolls horizontally on to a ramp that has a total height of 0.8m. what minimum velocity will allow the car to reach the top of the ramp/
1/2mv^2=mgh
The masses are the same, so they cancel out leaving the above relationship:
1/2v^2=gh
Solve for v:
v=sqrt*(2gh)
where
g=9.8m/s^2
and
h=0.8m
To calculate the minimum velocity required for the toy car to reach the top of the ramp, we can use the principle of conservation of energy.
The potential energy (PE) at the top of the ramp is equal to the kinetic energy (KE) of the car.
The potential energy is given by the formula: PE = m * g * h
where m is the mass of the car, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height of the ramp (0.8m).
The kinetic energy is given by the formula: KE = 0.5 * m * v^2
where v is the velocity of the car.
Setting the potential energy equal to the kinetic energy, we can solve for the minimum velocity:
m * g * h = 0.5 * m * v^2
Canceling out the mass (m) on both sides of the equation, we get:
g * h = 0.5 * v^2
Rearranging the equation to 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, we have:
v = √(2 * 9.8 * 0.8)
v = √15.68
v ≈ 3.96 m/s
Therefore, the minimum velocity required for the toy car to reach the top of the ramp is approximately 3.96 m/s.