What is the acceleration of the crate

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SUMMARY

The acceleration of a 100 kg crate on a 40-degree incline with a kinetic friction coefficient (μk) of 0.25 is calculated using the equation of motion. The correct approach involves accounting for both gravitational force and frictional force acting on the crate. The final calculation yields an acceleration of approximately -3.84 m/s², indicating the crate is accelerating down the incline. The error in the initial calculation was due to neglecting the cosine component of the frictional force.

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  • Understanding of Newton's second law of motion
  • Knowledge of static and kinetic friction coefficients
  • Basic trigonometry for resolving forces on an incline
  • Familiarity with free-body diagrams
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  • Learn about static vs. kinetic friction in detail
  • Explore the concept of free-body diagrams in physics
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Students studying physics, particularly those focusing on mechanics and dynamics, as well as educators looking for examples of incline problems in force analysis.

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Homework Statement


A 100 kg crate is carefully placed with zero velocity on the incline. What is the
acceleration of the crate if theta= 40degrees? μk=0.25 and μs=0.35(between the box and the incline.


Homework Equations





The Attempt at a Solution


My assumption is that,the (coefficient of)static friction will have no effect because we are looking at when the box starts sliding down the incline .Having that in mind,

i came up with: ∑F=m*a:
μk-Fg*sin40=m*a
245-630=100*a
a=-3.84 *******but this is wrong. Thanks in advance
 
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Resolved by moi.I forgot to multiply the friction by cos40:smile:
 

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