Force and acceleration on a waterslide

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Homework Help Overview

The discussion revolves around calculating the force exerted by a person of a certain mass while descending a waterslide, specifically the "Killimanjaro," which has a height of 49 meters and a maximum speed of 25.28 m/s. The original poster is considering various factors, including drag force, in their analysis.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to calculate acceleration using the "suvat" equation and considers the impact of drag force on their calculations. Participants question the clarity of the original poster's inquiry regarding the type of force being discussed and seek clarification on the direction of the force exerted on the slide.

Discussion Status

The discussion is actively exploring the nature of the force in question, with participants providing clarifications and prompting the original poster to specify their inquiry further. Some guidance has been offered regarding the relationship between the forces involved, particularly in relation to gravity and friction.

Contextual Notes

Participants note the importance of the slide's slope in determining the force applied due to gravity, indicating that additional information may be necessary for a complete analysis.

Alex no Sekai

Homework Statement


I am investigating how much force (F) a person of a certain mass (m) would need to go down one of the world's tallest waterslides, "Killimanjaro" for an important maths project.

Waterslide height: 49 m
Maximum speed: 25.28 m/s

Homework Equations


I feel like I am not taking enough factors into account. Should I look at the water's drag force, using the equation:

R = 0.5 p C A v^2
R = drag force
p = density
C = coefficient of drag
A = acceleration
v = velocity

The Attempt at a Solution


Photo le 07.07.17 à 13.22.jpg


I am sorry that this is messy however, I tried to find the acceleration by rearranging the "suvat" equation v=u+at. I got 6.52 ms^-2.

Next, I placed my value for a (6.52) in the equaiton F=ma (force = mass x acceleration). So that I can find any force, I just need to add the mass of the person within the equation. I think that I need to also take the drag force from the water into account though.
 
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Alex no Sekai said:
how much force (F) a person of a certain mass (m) would need

What exactly do you mean by this?
 
person123 said:
What exactly do you mean by this?
I am sorry that I was unclear. I am trying to see how much force a person would exert whilst going down the slide, taking into account their mass.
 
Alex no Sekai said:
how much force a person would exert

How much force a person would exert or the net force exerted on the person?
 
person123 said:
How much force a person would exert or the net force exerted on the person?
How much force a person would exert is what I am looking for.
 
Alex no Sekai said:
How much force a person would exert is what I am looking for.
Exert on what?
 
person123 said:
Exert on what?
Oh I forgot to mention that, sorry! It would be how much force that person exerts on the slide.
 
Alex no Sekai said:
Oh I forgot to mention that, sorry! It would be how much force that person exerts on the slide.

In which direction—normal to the slide?
 
person123 said:
In which direction—normal to the slide?
Yes. Thank you for trying to help me!
 
  • #10
Alex no Sekai said:
Yes. Thank you for trying to help me!

All right. In that case, the person's final velocity wouldn't be much help because that's affected by friction, which acts parallel to the slide.

It's very easy to find the force applied on the slide due to gravity as long as you have the slide's slope.
 
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  • #11
person123 said:
All right. In that case, the person's final velocity wouldn't be much help because that's affected by friction, which acts parallel to the slide.

It's very easy to find the force applied on the slide due to gravity as long as you have the slide's slope.
Thank you very much, this problem makes much more sense now!
 

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