Answer to Question #93917 in Mechanics | Relativity for Smayan

Question #93917
A child of mass 40kg at A climbs to the top of a playground slide B and slides down to C.

At the top of the slide the child has gained 1200J of potential energy .55% of this energy is lost to the surroundings in sliding down the 12 m from B to C (g=10N/kg)

Q calculate the mean force actibg against the motion of the child between B to C
1
Expert's answer
2019-09-09T10:53:20-0400

The work against friction:

Wfr=0.55EW_{fr}=0.55E

The mean force acting against the motion of the child between B to C:


F=Wfrs=0.55EsF=\frac{W_{fr}}{s}=\frac{0.55E}{s}

F=0.55(1200)12=55 NF=\frac{0.55(1200)}{12}=55\ N


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