A 42-kg child slides from the top of a 3.0m high playground slide. With that speed will the child reach the bottom of the slide assuming: a.) there is no friction? b.) 23 J energy is dissipated due to friction?

1 Answer
LM
Feb 3, 2018

(a) #5.48 m//s# (b) #5.43 m//s#

Explanation:

#KE = mv^2#

#v^2 = (KE)/m#

#v = sqrt((KE)/(m))#

where

#v# is velocity in metres per second

#KE# is kinetic energy in joules

and #m# is mass in kilograms.

#m = 42kg#

#GPE = mgh#

#= 42kg * 10 N//kg * 3m#

#= 1260 J#

#GPE#, gravitational potential energy, is lost when the child goes down the slide.
the #GPE# lost is equal to the #KE#, kinetic energy gained - assuming there is no friction.

#KE = 1260 J#

#v = sqrt((KE)/(m)) = sqrt((1260)/(42))#

#= sqrt(30)#

#v = sqrt(30) = 5.48 m//s# (3s.f.)

without friction, the speed of the child reaching the bottom is #5.48 m//s#.

if #23J# of energy is dissipated due to friction, then there are #23J# less of #KE#.

#1260J - 23J = 1237 J#

the mass of the child stays the same.

#v = sqrt((KE)/(m)) = sqrt((1237)/(42))#

#=5.43 m//s#

with #23J# dissipated due to friction, the speed of the child reaching the bottom is #5.43 m//s#.

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