TP 7.56

Given:

m = 0.160 kg 

V₀ = 0 m/s

y₀ = 3.50 m

(a) The power supplied by gravity is given by the product of weight (which stays constant) and speed of the apple (which increases). Since the speed increases, so does the power.

(b) P - ? at  y = 2.50 m

P = F V = (mg) V

V = (V₀² - 2g (y - y₀))^1/2  

V = (0 - 2(9.81 m/s²) ((3.00 m) - (2.50 m)))^1/2  = 3.13 m/s

P =   (0.160 kg) (9.81 m/s²) (3.13 m/s) = 4.91 W

(c) P - ? at  y = 1.50 m

P = F V = (mg) V

V = (V₀² - 2g (y - y₀))^1/2  

V = (0 - 2(9.81 m/s²) ((3.00 m) - (1.50 m)))^1/2  = 5.42 m/s

P =   (0.160 kg) (9.81 m/s²) (5.42 m/ s) = 8.51 W