MHB Work energy principle and power

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The discussion revolves around calculating the loss of gravitational potential energy (GPE) and the distance traveled by a child sliding down a playground slide. The child, with a mass of 40kg, experiences a loss of GPE of 800J when descending from a height of 2m. With a constant resistance of 112N, the distance traveled before coming to rest is calculated to be 7.14m. For the level part of the slide, the increase in GPE is corrected to -400J, leading to a distance of 4m traveled on the level section. The final distance on the level part is determined to be 3.14m after subtracting the distances.
Shah 72
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A child of mass 40kg slides down a playground slide. The child starts from rest at the top of the slide, 2m above the ground. At the bottom of the slide it's slope levels off.
a) Find the child's loss of GPE
I got the ans 800J
there is a constant resistance of 112N throughout
b) find the distance the child has traveled when she comes to rest.
Using work energy principle
Increase in KE =0J
Increase in GPE= -800J
Work done against resistance = -112SJ
So I get S= 7.14m
The slide is inclined at an angle of 30 degree to the horizontal.
C) Find the distance the child travels on the level part of the slide.
I don't understand this part.
Increase in KE= 0J
Increase in GPE= 0- 40×10sin30×2= -400J
Work done by gravity= F×S
I don't understand how to solve further. Pls help
 
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Shah 72 said:
A child of mass 40kg slides down a playground slide. The child starts from rest at the top of the slide, 2m above the ground. At the bottom of the slide it's slope levels off.
a) Find the child's loss of GPE
I got the ans 800J
there is a constant resistance of 112N throughout
b) find the distance the child has traveled when she comes to rest.
Using work energy principle
Increase in KE =0J
Increase in GPE= -800J
Work done against resistance = -112SJ
So I get S= 7.14m
The slide is inclined at an angle of 30 degree to the horizontal.
C) Find the distance the child travels on the level part of the slide.
I don't understand this part.
Increase in KE= 0J
Increase in GPE= 0- 40×10sin30×2= -400J
Work done by gravity= F×S
I don't understand how to solve further. Pls help
I think I did mistake while calculating Increase in GPE, it will be -800J
Work done by gravity = 200 SJ
So I get S= 4m
I subtract from full length 7.14-4= 3.14m
 

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