What minimum speed does a 100 g puck need to make it to the top of a 5.9 m-long, 29 degree frictionless ramp?
To rise to the top of the ramp, the initial kinetic energy must be at least
M*g*5.9 sin29, the potential energy there.
(1/2)M Vo^2 = M*g*(5.9 sin29)
M cancels out.
Vo = sqrt(2*g*5.9sin29) = 7.5 m/s
To determine the minimum speed required for the 100 g puck to make it to the top of the 5.9 m-long, 29 degree frictionless ramp, we can use the principles of energy conservation.
First, let's consider the initial and final positions of the puck. The initial position is at the bottom of the ramp, and the final position is at the top of the ramp. We can assume that the initial velocity of the puck is zero.
The potential energy at the bottom of the ramp can be calculated using the mass (100 g) and the height difference between the bottom and top of the ramp. In this case, the height difference is given by:
Δh = (5.9 m) * sin(29 degrees)
The potential energy at the top of the ramp is zero since the puck has reached its highest point.
Using the principle of conservation of energy, we can equate the initial potential energy to the final kinetic energy. Since the initial kinetic energy is zero, we can focus only on the potential energy.
Therefore, we have the equation:
m * g * Δh = 1/2 * m * v^2
Where:
m = mass of the puck (100 g = 0.1 kg)
g = acceleration due to gravity (approx. 9.8 m/s^2)
Δh = height difference between the bottom and top of the ramp (5.9 m * sin(29 degrees))
v = velocity of the puck at the top of the ramp
Simplifying the equation, we get:
g * Δh = 1/2 * v^2
Now we can solve for v by rearranging the equation:
v = √(2 * g * Δh)
Plugging in the values:
v = √(2 * 9.8 m/s^2 * 5.9 m * sin(29 degrees))
Evaluating the expression, we find:
v ≈ 7.84 m/s
Therefore, the puck needs to have a minimum speed of approximately 7.84 m/s to reach the top of the ramp.