Santa loses his footing and slides down a frictionless, snowy roof that is tilted at an angle of 26.0. If Santa slides 7.00 before reaching the edge, what is his speed as he leaves the roof?
vertical distance fell: 7sin26
1/2 m v^2=mgh
v= sqrt 2gh
10.23
To determine Santa's speed as he leaves the roof, we can use the principle of conservation of mechanical energy. Since the roof is frictionless, the only forces acting on Santa are gravity and the normal force. The normal force can be decomposed into its components perpendicular and parallel to the roof.
First, let's determine the height (h) of the roof using the given angle (θ) and the distance (d) Santa slides before reaching the edge:
h = d * sin(θ)
h = 7.00 * sin(26.0°)
h ≈ 3.106 meters
Now, let's calculate Santa's speed (v) as he leaves the roof using the principle of conservation of mechanical energy. At the start of his slide on the roof, Santa has potential energy equal to his weight times the height, and no kinetic energy:
Potential energy at the start = m * g * h
Kinetic energy at the end = (1/2) * m * v^2
Since energy is conserved, we can equate the two:
m * g * h = (1/2) * m * v^2
Simplifying by canceling out the mass (m), we have:
g * h = (1/2) * v^2
Rearranging the equation to solve for Santa's speed (v), we get:
v = √(2 * g * h)
Plugging in the values, where g is the acceleration due to gravity (approximately 9.8 m/s^2):
v ≈ √(2 * 9.8 * 3.106)
v ≈ √(60.688)
v ≈ 7.8 meters/sec
Therefore, Santa's speed as he leaves the roof is approximately 7.8 meters per second.