Rates of Power before and after

Question

the component of the weight parallel to the surface of an inclined plane is mgsin() where  is the angle of the incline with respect to the horizontal direction. A 2.0-kg box slides downward along a frictionless inclined plane that makes an angle of 15° with the horizontal direction. The initial speed of the box equals 3.0 m/s. At what rates of power, P, does the gravitational force do work on the box initially, and after the block has slid 4.0 m from its initial position (if one may take for the value of the gravitational acceleration, g = 9.8 m/s2 downward and if one may neglect any effects due to air resistance)?

Answer 1

Power = Rate of Doing work = Force*distance /time = Force * velocity

Initial velocity = 3m/s

Force = mgSin(theta ) = 2*9.8*Sin 15 = 5.07 N

Power at start = 5.07* 3 = 15.22W

Velocity after 4m

V^{2 =} U² +2gSin (theta) *Distance = 3² + 2*9.8* Sin 15 = 9 + 5.07 =20.07

V=4.48m/s

Thus Power after sliding = Force * Velocity =( mgSin 15) 4.48 = 22.73 W