Answer:
vf = 12.343 m/s
Explanation:
Given:
- The Length of the ramp s = 30 m
- The angle of the ramp θ = 15 degrees
- Initial velocity at the top of ramp vi = 0
- Neglect Friction
Find:
How fast is Albert Jr. going at the bottom of the ramp?
Solution:
- We can use conservation of energy for the skateboard at top and bottom of the ramp.
ΔK.E = ΔP.E
Where, ΔK.E : Change in kinetic energy
ΔP.E : Change in gravitational potential energy
0.5*m*(vf^2 - vi^2) = m*g*Δh
Where, m: The mass of the object
Δh: Change in elevation from top to bottom
Δh = s*sin(θ)
- Substitute and simplify:
0.5*(vf^2 - vi^2) = g*s*sin(θ)
vf^2 = 2*g*s*sin(θ)
vf = √(2*g*s*sin(θ)) = √(2*9.81*30*sin(15))
vf = 12.343 m/s
- The velocity at the bottom of ramp would be 12.343 m/s
