What minimum speed does a 170 g. puck need to make it to the top of a 6.0 m. long, 27 frictionless ramp?
its 27 degree horizontal **
The vertical rise of the puck is
H = 6.0 sin27 = 2.724 m
Conservation of energy tells you that, when there is no friction, and the velocity is zero at the top,
gH = Vo^2/2
Solve for the required minimum initial speed Vo.
You don't need to know the mass.
To determine the minimum speed required for a 170 g puck to make it to the top of a 6.0 m long, frictionless ramp, we can use the principle of conservation of energy.
The potential energy at the top of the ramp should be equal to the initial kinetic energy of the puck. The formula for potential energy is given by:
PE = m * g * h
where PE is the potential energy, m is the mass of the puck (in kg), g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height.
Since we want to find the minimum speed required to reach the top of the ramp, we can write the equation as:
PE = (1/2) * m * v^2
where v is the velocity of the puck.
Given that the mass of the puck (m) is 170 g or 0.17 kg, the height (h) is 6.0 m, and the acceleration due to gravity (g) is 9.8 m/s^2, we can write the equation as:
(1/2) * 0.17 kg * v^2 = 0.17 kg * 9.8 m/s^2 * 6.0 m
Simplifying the equation:
0.5 v^2 = 9.8 * 0.17 * 6
Multiply the values:
0.5 v^2 = 9.8 * 0.17 * 6
0.5 v^2 = 9.882
v^2 = 9.882 / 0.5
v^2 = 19.764
Taking the square root of both sides:
v = √19.764
v ≈ 4.44 m/s
Therefore, the minimum speed required for the 170 g puck to make it to the top of the 6.0 m long, frictionless ramp is approximately 4.44 m/s.