4. A 25 kg boy started from rest and slid from the top of a water slide. The fri

Question

4. A 25 kg boy started from rest and slid from the top of a water slide. The friction force between the boy and the water slide was 10 N. g 9.8 m/s a. What would the boy's velocity be at the bottom of the water slide (using constant acceleration method)? b. What would the boy's velocity be at the bottom of the water slide (using work-energy method)? c. The parents of the boy think the final velocity is too fast. What changes can be made to help the boy reach a slower final velocity? 6 points 10m 5m 30 deg

Explanation / Answer

4. (A) F_net = m g sin(30) - f

F_net = (25 x 9.8 x sin30) - 10 = 112.5 N  

a = F/m = 112.5 / 25 = 4.5 m/s^2

vf^2 - vi^2 = 2 a d

v^2 - 0 = 2(4.5)(10)

v = 9.5 m/s

(B) Work done by gravity + work done by friction = change in KE

m g d sin30 - f d = m v^2 /2 - 0

v = sqrt[ (2 x 9.8 x 10 sin39) - (2 x 10 x 10 / 25)]

v = 9.5 m/s

(C) to reduce speed,

increase friction force.

reduce the length

reduce the angle

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