Rabu, 08 Maret 2017

a child sleds down a frictionless hill with vertical drop h. at the bottom is a level stretch where the coefficient of friction is 0.27.

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Answers

Best Answer:  (mu)(mass)(gravity)(distance)=1/2m(veloc...
mass cancels
(mu)(g)(d)=1/2v^2
sqrt((0.27)(9.8)(19)*2)=v
v=10.03 m/s

that means her initial velocity as she left the slide was 10.03 m/s


v^2=2gh
(10.03)^2=2(9.8)h
h= 5.13 m

The friction force causes her velocity to decrease from her velocity to 0 m/s at she slides 19 meters.


Ff = 027 * m * 9.8 = m * 2.646
a = -2.646 m/s^2

Let’s use the following equation to determine her velocity at the bottom of the hill.


vf^2 = vi^2 + 2 * a * d, vi = 0
0 = vi^2 + 2 * -2.646 * 19
vi^2 = 100.548
vi = √100.548

This is approximately 10.027 m/s. Since the hill is frictionless, we can use conservation of potential and kinetic energy to determine its height. Let’s determine her kinetic energy at the bottom of the hill.

KE = ½ * m * 100.548 = m * 50.274
PE = m * 9.8 * h
m * 9.8 * h = m * 50.274
h = 50.274 ÷ 9.8 = 5.13 meters

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a child sleds down a frictionless hill with vertical drop h. at the bottom is a level stretch where the coefficient of friction is 0.27.
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