Acceleration of Book on Incline: Forces & Friction Analysis

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SUMMARY

The discussion focuses on calculating the acceleration of a 3.91 kg physics book on a 21.7° incline, connected to a 519.0 g coffee cup. The coefficients of static and kinetic friction are μs = 0.457 and μk = 0.330, respectively. The user attempts to solve the problem by analyzing the forces acting on both the book and the coffee cup, but encounters difficulties in deriving the correct tension and acceleration values. The key calculation involves determining the weight components and frictional forces acting on the book.

PREREQUISITES
  • Understanding of Newton's second law (F_N_E_T=ma)
  • Knowledge of forces on inclined planes
  • Familiarity with static and kinetic friction coefficients
  • Ability to resolve forces into components
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  • Review the derivation of tension in systems involving pulleys and inclined planes
  • Study the calculation of frictional forces using coefficients of friction
  • Learn about free-body diagrams for multi-object systems
  • Explore advanced problems involving acceleration on inclined planes
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Students studying physics, particularly those focusing on mechanics and forces, as well as educators seeking to clarify concepts related to inclined planes and friction.

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Homework Statement



The 3.91 kg physics book shown is connected by a string to a 519.0 g coffee cup. The book is given a push up the slope and released with a speed of 3.99 m/s. The coefficients of friction are μs = 0.457 and μk = 0.330. What is the acceleration of the book if the slope is inclined at 21.7°?

Homework Equations



[tex]F_N_E_T=ma[/tex]

The Attempt at a Solution



I've tried splitting the equation into two: the book and the coffee mug.

COFFEE MUG:

[tex]F_N_E_T=\vec{F}_g - \vec{F}_T[/tex]
[tex]ma=mg - T[/tex]
[tex]0.519a=(0.519)(9.8)-T[/tex]
[tex]T=5.0862-0.519a[/tex]

BOOK:

I've split weight into its x and y components.

[tex]F_N_E_T = T + W_x - F_k_i_n_e_t_i_c[/tex]
[tex]ma = T + W_x - F_k_i_n_e_t_i_c[/tex]
[tex]3.91a = T + 8.152[/tex]
[tex]T = 3.91a - 8.152[/tex]

Then I equated T and solved for a. But I'm still getting the wrong answer...
 
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can you describe how you got the 8.152N?
 

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