How to calculate the change in kinetic energy on a plane?

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SUMMARY

The discussion focuses on calculating the change in kinetic energy of a crate being pulled up a rough incline. The crate, with a mass of 11 kg and an initial velocity of 1.55 m/s, is subjected to a 150 N force over a distance of 9.33 m at an incline of 20°. The correct approach involves accounting for both the frictional force and the gravitational component acting parallel to the incline. The final calculation for work done, which equals the change in kinetic energy, is determined to be 1399.5 J after correcting for the gravitational force component.

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  • Understanding of Newton's laws of motion
  • Familiarity with the concepts of kinetic energy and work-energy principle
  • Knowledge of frictional forces and their calculations
  • Basic trigonometry for resolving forces on an incline
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  • Explore the effects of friction on motion in physics
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governergrimm
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Homework Statement


A crate is pulled by a force (parallel to the incline) up a rough incline. The crate has an initial speed shown in the figure below. The crate is pulled a distance of 9.33 m on the incline by a 150 N force. The acceleration of gravity is 9.8 m/s 2 . What is the change in kinetic energy of the crate? Answer in units of J.
Mass = 11kg
Angle of the incline (θ)= 20°
Initial Velocity = 1.55 m/s
μ = .308
D= 9.33 Meters
Force Applied = 150N
g=9.8

Homework Equations


Ff= μN
W=Fd=ΔKE
Fapplied -Ff=Fnetx
FnetVertical =0

The Attempt at a Solution


N-mgcosθ=0
N=mgcosθ
Fa-Ff=Fnet
Fa-μN
Fa-μmgcosθ=Fnet
150-(.308)(11)(9.8)cos(20°)=118.8
W=ΔKE = 118.8(9.33) = 1399.5

Not sure what I am missing but it is incorrect.
 
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You've forgotten the other (parallel-to-the-plane) component of gravity.
 
My force should look like this then?

force applied - force friction - mgsinθ= Fnet

Is that all that I'm missing?
Thanks!
 

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