Final answer:
The force of gravity acting down a 15 m ramp inclined at 30° on a 15 kg block is approximately 73.5 N. The normal force is approximately 127.4 N. The block will accelerate at about 4.9 m/s² and reach a final speed of about 17.1 m/s at the bottom of the ramp, assuming there is no friction.
Step-by-step explanation:
The problem involves a 15 kg block on top of a 15 m ramp inclined at 30°.
- a. The force of gravity acting down the ramp (component of gravity parallel to the incline) can be found using the formula:
Fg = m * g * sin(θ), where m = 15 kg, g = 9.8 m/s², and θ = 30°. After calculations, Fg is approximately 73.5 N. - b. The normal force is the component of gravity perpendicular to the incline. It can be calculated using: Fn = m * g * cos(θ). Hence, Fn is approximately 127.4 N.
- c. The block's acceleration down the ramp due to gravity, a, can be found using a = g * sin(θ), which computes to approximately 4.9 m/s².
- d. The final speed of the block at the bottom can be calculated using the kinematic equation v = sqrt(2 * a * d), where d is the length of the ramp. This results in a final speed of approximately 17.1 m/s, assuming no friction.