Answer to Question #93917 in Mechanics | Relativity for Smayan

Question

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

Answer 1

The work against friction:

$$W_{fr}=0.55E$$

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

$$F=\frac{W_{fr}}{s}=\frac{0.55E}{s}$$

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