A 3.2 kg crates slides along a 6.0 m horizontal surface with speed of 5.0m/s. It then moves down along an incline with a lenth 6.0m and a height from the ground of 4.5 mand then along a second, lower horizontal surface. The second horizontal surface has a frictional force of 3.0 N. Assuming all surfaces are frictionless, find the:
A) total mechanical points at point A and B
B) speed when the crate arrives at point B
A) Total mechanical energy at point A: KE + PE = (1/2)mv^2 + mgh = (1/2)(3.2 kg)(5.0 m/s)^2 + (3.2 kg)(9.8 m/s^2)(4.5 m) = 97.6 J
Total mechanical energy at point B: KE + PE = (1/2)mv^2 + mgh = (1/2)(3.2 kg)(v^2) + (3.2 kg)(9.8 m/s^2)(0 m) = (1/2)(3.2 kg)(v^2)
B) Speed when the crate arrives at point B: Using the conservation of energy, we can set the total mechanical energy at point A equal to the total mechanical energy at point B.
97.6 J = (1/2)(3.2 kg)(v^2)
v = √(97.6 J / (1/2)(3.2 kg)) = 8.2 m/s