Crate pulled up incline, kinetic energy and speed; picture included

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SUMMARY

The discussion focuses on the physics problem of calculating the change in kinetic energy and final speed of a crate being pulled up a rough incline by a 150 N force over a distance of 5.94 m. The acceleration due to gravity is given as 9.8 m/s². The participant attempted to use the work-energy principle, applying the equation W = ΔK, and calculated the work done against friction to arrive at a change in kinetic energy of 147.63 J. Subsequently, they used this value to compute the final speed of the crate, resulting in approximately 4.77 m/s.

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  • Understanding of Newton's laws of motion
  • Familiarity with the work-energy theorem
  • Knowledge of kinetic energy calculations
  • Basic grasp of forces acting on an incline
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  • Learn about frictional forces and their impact on motion
  • Explore the concept of conservation of energy in mechanical systems
  • Practice problems involving forces on inclined planes
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Students studying physics, particularly those focusing on mechanics and energy concepts, as well as educators looking for practical examples of kinetic energy and force applications in real-world scenarios.

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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 5.94 m on the incline by a 150 N force.
The acceleration of gravity is 9.8 m/s2 .
[PLAIN]http://img801.imageshack.us/img801/849/problem1011.jpg
a) What is the change in kinetic energy of the crate?
Answer in units of J.
b) What is the speed of the crate after it is pulled the 5.94 m?
Answer in units of m/s.

Homework Equations


(1)W= delta K= Integral Fx dx= Integral Sum of F dr= Integral Sum of F dr + Integral fk dr= Sum of W - fkd
(2)Delta K= 1/2 mvf^2

The Attempt at a Solution


Wells for part a) I tried equation (1)and got 150 N- (0.399* 5.94)=147.63 J... which I don't know if its right
and for b) (if part a is right) I used equation (2) vf=sq rt ( (2/m)*147.63)= 4.76574 m/s
 
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When using W =delta K , W includes the work done by gravity or any other conservative forces, when present. It is best to use conservation of total energy, Work done by non - conservative forces (like friction and the applied force) equals delta __??__ + delta __??_.

The problem asks for the change in Kinetic Energy, in addition to its speed.

You are getting your energy and force units mixed up.
 

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