Calculating Energy and Friction in a Child's Descent on a 6m Slide

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SUMMARY

The discussion focuses on calculating the energy lost to friction during a child's descent on a 6m slide, specifically analyzing a scenario where a 26kg child descends 2m vertically. The potential energy at the top is calculated as 509.6J (using the formula PE = mgh), while the kinetic energy at the bottom is 208J (using KE = 0.5mv²). The energy lost to friction is determined by the difference between these two values, indicating that 301.6J is the energy used to overcome friction during the descent.

PREREQUISITES
  • Understanding of potential energy (PE = mgh)
  • Knowledge of kinetic energy (KE = 0.5mv²)
  • Familiarity with the concept of energy conservation
  • Basic principles of friction and work done (W = Force × displacement)
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  • Study the principles of energy conservation in mechanical systems
  • Learn about calculating work done against friction
  • Explore the effects of different slide angles on frictional forces
  • Investigate real-world applications of energy loss in motion
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A child of mass 26kg is traveling at 4.0ms at the bottom of the slide which is 2m vertically below the top. The length of the slide is 6m.
Calculate

i) The energy used in overcoming the friction during the child's descent

ii) The magnitude of the frictional force.

Totally confused about this question. I know Work done = Force.displacement in the direction of the force but then does this mean that question i) is equal to W=26(9.8)Cos19.47?

I have no idea. Would appreciate any help.

thank you
 
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Assume the child started at the top with zero speed. What's the total energy at the top? At the bottom?
 
At the top would be the potential energy so 26*9.8*2, but this would be the energy output (I think). That's not really what i) is asking. At the bottom would also be the same but in kinetic energy form.
 
ZAMM said:
At the bottom would also be the same but in kinetic energy form.
Why don't you check and see? They give you the speed. (The energy would only be the same if there were no losses to friction.)
 
OK, so at the bottom of the slide E=.5(26)(4)^2 but I'm still confused. The length of the slide has 2 components? is the horizontal component the frictional component? I have no clue.
 
Potential energy only deals with vertical displacement. You need the length of the slide for the second part of the question
 
I know it only deals with vertical displacement but in this context the energy that requires him to get from the top to the bottom (neglecting the slide) would be mgh, but this alone does not help me solve the equation.
 
ZAMM said:
OK, so at the bottom of the slide E=.5(26)(4)^2 but I'm still confused.
Good. How does that compare to the energy it started with at the top?
 
The problem deals with conservation of energy. Etop=Ebottom. Without friction, PEtop=KEbottom, but that's not what you have here.
 
  • #10
At the top PE=mgh=26*9.8*2=509.6J and at the bottom KE=.5*26*4^2=208J. Energy is greater at the top than at the bottom.
 
  • #11
OK, so what's the answer to the first question?
 
  • #12
The difference between them? but if it is that, what I calculated is not energy "used" but more wasted.
 
  • #13
ZAMM said:
The difference between them? but if it is that, what I calculated is not energy "used" but more wasted.
That depends on your point of view: You "used" it to overcome friction.
 

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