Penguin on a slide with friction: find the angle of the slide

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Homework Help Overview

The problem involves a penguin sliding down a slide with friction, where the coefficient of kinetic friction is given. The challenge is to find the angle of the slide based on the time taken to slide down compared to a frictionless scenario.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss using mechanical energy equations and equations of motion to analyze the problem. There are questions about the appropriateness of using energy conservation for this scenario, with suggestions to derive time equations for both frictional and frictionless cases.

Discussion Status

Some participants have provided guidance on using equations of motion and drawing free body diagrams (FBD) to identify forces and accelerations. There is an ongoing exploration of different approaches, with no explicit consensus on the method to be used.

Contextual Notes

Participants are working under the constraint of comparing the time taken to slide down the incline with and without friction, which is central to the problem's requirements.

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1.Question:

A penguin is going down a slide. The coefficient of kinetic friction between the penguin and the slide has a value of 0.77. It takes three times as long to slide down this slide as it would if the slide were frictionless. Find the angle of the slide.
2. My answer: I chose solving it using mechanical energies

Suppose it reaches ground state after covering distance d,

Equations of mechanical energy;

with friction:

Ek+Epg=Wf+Ek1+Epg1 Wf=-0.77.d.mg.cos(alfa)

mv^2/2+mg.d.sin(alfa)=-0.77dmgcos(alfa)+mV^2/2+mgh(final) divide by m (h final=0 taking ground state when it covers d)

v^2/2+gdsin(alfa)=-0.77gdcos(alfa)+V^2/2--------------------{1}

without friction: Ek+Epg=Ek1+Epg1 (Epg1=0)

mv^2/2+mg.3dsin(alfa)=mV^2/2 divide by m (V after cutting 3d without friction equal V after cutting 1d with friction)

v^2/2+3d.g.sin(alfa)=V^2/2----------------------------{2}

{2} - {1}:... tan(alfa)=0.77.g/2

that gives ALFA=75.155 degree Is it correct?
 
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Any Help said:
Ek+Epg=Wf+Ek1+Epg1 Wf=-0.77.d.mg.cos(alfa)

mv^2/2+mg.d.sin(alfa)=-0.77dmgcos(alfa)+mV^2/2+mgh(final) divide by m (h final=0 taking ground state when it covers d)

v^2/2+gdsin(alfa)=-0.77gdcos(alfa)+V^2/2--------------------{1}

without friction: Ek+Epg=Ek1+Epg1 (Epg1=0)

mv^2/2+mg.3dsin(alfa)=mV^2/2 divide by m (V after cutting 3d without friction equal V after cutting 1d with friction)

v^2/2+3d.g.sin(alfa)=V^2/2----------------------------{2}

{2} - {1}:... tan(alfa)=0.77.g/2

that gives ALFA=75.155 degree Is it correct?

There are multiple mistakes in your solution, Are you asked to solve this using energy conservation?
You can basically use equations of motion to solve this
It gave you an equation.
## 3T_{time ~takes~ to ~slide ~without ~friction} = T_{time~ takes ~to ~slide~ with ~friction} ##
Draw a FBD, identify the forces and acquire accelerations of the bodies.

Some algebra will give you the angle, Everytime you arrive at something post your work and we will be here :D
 
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Biker said:
There are multiple mistakes in your solution, Are you asked to solve this using energy conservation?
You can basically use equations of motion to solve this
It gave you an equation.
## 3T_{time ~takes~ to ~slide ~without ~friction} = T_{time~ takes ~to ~slide~ with ~friction} ##
Draw a FBD, identify the forces and acquire accelerations of the bodies.

Some algebra will give you the angle, Everytime you arrive at something post your work and we will be here :D
you mean i must find the time equation for both?
 
Any Help said:
you mean i must find the time equation for both?
Yes :D
Use the equations of motion to find the time.
 
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then it will be with friction
-0.77gcos(alfa)+gsin(alfa)=af
without friction gsin(alfa)=a
delta x= v0t+0.5at^2
t=sqaur root (2.deltax/af) T=sqrrt(2.deltax/a)
3T=t
9T^2=t^2
9af=a
tan(alfa)=9x0.77/8 alfa=40.9
right?
 
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Any Help said:
then it will be with friction
-0.77gcos(alfa)+gsin(alfa)=af
without friction gsin(alfa)=a
delta x= v0t+0.5at^2
t=sqaur root (2.deltax/af) T=sqrrt(2.deltax/a)
3T=t
9T^2=t^2
9af=a
tan(alfa)=9x0.77/8 alfa=40.9
right?
Excellent work :).
 
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