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A block with a mass of 1kg starts to slide down along the inclined plane with an inclination of 37 ° and the inclined plane is long enough (1) If the slope is smooth, the work done by gravity within 4 s of the block falling is calculated; (2) The dynamic friction factor between the object and the inclined plane is 0.1. Calculate the average power of gravity work within 4 s of the object falling, the instantaneous power of gravity work at the end of 4 S, and the instantaneous power of friction work at the end of 4 s
(1) The gravity of an object is divided into the component forces along the inclined plane and the vertical inclined plane,
The force along the inclined plane is f = mgsin37 °, the acceleration is f / M = gsin37 °, the displacement is T ^ 2gsin37 ° / 2 = 8gsin37 ° and the work done is FS = 8m (gsin37 °) ^ 2 = 8 (gsin37 °) ^ 2
(2) 1. The work done by gravity is constant, and the average power is FS / 4 = 2m (gsin37 °) ^ 2,
2. The friction force along the inclined plane is f = 0.1mgcos 37 ° and the acceleration is (f-f) / M = GSIN 37 ° - 0.1gcos 37 °,
The instantaneous power of gravity work is FV = fat = 4 (gsin37 °) * (gsin37 ° - 0.1gcos37 °)
The instantaneous power of friction: FV = fat = 0.4 (gcos37 °) * (gsin37 ° - 0.1gcos37 °)
Isn't it detailed yet? There are formulas and data
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