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

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SUMMARY

The discussion focuses on calculating the angle of a slide using the coefficient of kinetic friction, specifically 0.77, and the relationship between sliding time with and without friction. The initial approach using mechanical energy conservation was corrected, emphasizing the use of equations of motion to derive the angle. The final calculated angle of the slide is approximately 40.9 degrees, determined through the relationship between acceleration with and without friction.

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  • Understanding of Newton's laws of motion
  • Familiarity with kinetic friction and its coefficient
  • Basic knowledge of energy conservation principles
  • Ability to solve algebraic equations
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  • Study the equations of motion in physics
  • Learn about the effects of friction on motion
  • Explore the derivation of acceleration in inclined planes
  • Investigate the relationship between time, distance, and acceleration
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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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