A Dynamics Problem Regarding Friction.

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SUMMARY

The discussion centers on a dynamics problem involving two masses and friction. The coefficient of static friction between the 5.0 kg mass and the horizontal surface is established as 0.33. The coefficient of static friction between the two masses is also discussed but not explicitly solved. The critical question regarding the time it takes for the top mass to fall off the bottom mass is resolved, with the answer being 0.97 seconds.

PREREQUISITES
  • Understanding of Newton's Second Law (F=ma)
  • Knowledge of static friction and its coefficient (μ)
  • Familiarity with kinematic equations, specifically Δd=viΔt+½aΔt²
  • Basic concepts of relative motion in physics
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  • Study the derivation and application of the coefficient of static friction in various scenarios
  • Learn about kinematic equations and their applications in solving motion problems
  • Explore relative motion concepts in physics to understand how different frames of reference affect calculations
  • Investigate dynamics problems involving multiple bodies and friction to enhance problem-solving skills
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Students studying physics, particularly those focusing on dynamics and friction, as well as educators looking for examples of problem-solving in mechanics.

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Homework Statement



SSRnLDS.png


That was the given information. Here are the questions they ask, I have solved a) and b), but I am completely stuck on c). I'll just put them here in case information from them are necessary.

a) What is the coefficient of static friction between the 5.0 kg mass and the horizontal surface? (answer 0.33)

b) What is the coefficient of static friction between the two masses?

c) From the point when the top mass starts sliding how long will it take for the top mass to fall off the bottom mass? (0.97s)

Homework Equations



F=ma
fric = μ|Fn|

The Attempt at a Solution



Block on Top (2.0kg)
ƩFx=-fk
ƩFx=-(0.3)(2kg*9.81m/s^2)
ma=-(0.3)(2kg*9.81m/s^2)
2kg*a=-5.886N
a= \frac{-5.886N}{2kg}
a= -2.943m/s^2

then...
Δd=viΔt+\frac{1}{2}aΔt^2
-1m=\frac{1}{2}2.943m/s^2*Δt^2
-1m=-1.4715m/s^2*Δt^2
-1m=-1.4715m/s^2*Δt^2
\frac{-2m}{-1.4715m/s^2}=Δt^2
Δt^2=0.68s^2
Δt=0.82s^Which is wrong =(
 
Last edited:
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You've found the 2kg's acceleration relative to the ground (except, not sure why you have it as negative; isn't positive to the right?). Is that the acceleration you need to answer the question?
 
So does that mean I need to find the acceleration relative to the bottom block?
 
AvocadosNumber said:
So does that mean I need to find the acceleration relative to the bottom block?
You want the time it takes the top block to move 1m, starting from 'rest'. 1m relative to what? "rest" relative to what?
 
Oh... I see now... relative to the bottom block.
 
Quite so.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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