Question

Problem 4 A child on a sled slides (starting from rest) down an icy slope that makes an angle of 15 with the horizontal. After sliding 20 m down the slope, the child enters a flat, slushy region, where she slides for 2.0s with a constant negative acceleration of -1.5 m/s with respect to her direction of motion. She then slides up another icy slope that makes a 20 angle with the horizontal. (a) How fast was the child going when she reached the bottom of the first sope? How long did it take he to get there? (b) How long was the flat stretch at the bottom? (c) How fast was the child going as she started up the second slope? (d) How far up the second slope did she slide?

Answer 1

a) mgl sin 150 = 1/2 mv^2 doublerightarrow v = squareroot 2 times 9.8 times 20 times sin 15 degree = 10.07 m/s 20 m = 1/2 at^2 doublerightarrow t = squareroot 2 times 20/9.8 times sin 15 degree doublerightarrow t = 3.97 s b) s = 10.07 times 2 - 1/2 times (1.5) times (2)^2 = 17.18 m c) V = u - at = 1.07 - 1.5 times 2 = 7.07 m/s d) mgl sin theta = 1/2 mv^2 rightarrow l = (7.07)^2/9.8 times sin 20 degree doublerightarrow l = 14.95 m