Calculating Change in Kinetic Energy for a Crate Pulled Up an Incline

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SUMMARY

The discussion focuses on calculating the change in kinetic energy for a crate being pulled up a rough incline by a force of 150 N over a distance of 9.17 m. The initial calculations attempted by users involved the forces acting on the crate, including gravitational and frictional forces, but initial results were incorrect. The correct change in kinetic energy was determined to be 533.0427 J. To find the final speed of the crate after being pulled, users were advised to first calculate acceleration using Newton's second law and then apply kinematic equations.

PREREQUISITES
  • Understanding of Newton's second law (F = ma)
  • Knowledge of kinematic equations for uniformly accelerated motion
  • Familiarity with concepts of kinetic energy (Ek = 0.5mv²)
  • Basic understanding of forces acting on an object on an incline, including friction
NEXT STEPS
  • Study the application of Newton's second law in inclined plane problems
  • Learn how to calculate acceleration from net forces acting on an object
  • Explore kinematic equations for solving motion problems, particularly Vf = Vi + at
  • Review the principles of energy conservation in mechanical systems
USEFUL FOR

Students and educators in physics, particularly those studying mechanics and energy concepts, as well as anyone involved in solving problems related to forces on inclined planes.

BallerRegis
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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 9.17 m on the
incline by a 150 N force.
The acceleration of gravity is 9.8 m/s2 .
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 9.17 m?
Answer in units of m/s

For part A i tried
F-mg[sin theta - (friction x cos theta)] and got 135.232 but got wrong answer then i tried multiplying it by the distance and got it wrong.
Also i tried using F - [ μ m g cos θ ] and got it wrong
Someone please help me
 
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The question appears to be incomplete. Do you know the angle, initial speed and coefficient of friction? If so, your calculation of
"F-mg[sin theta - (friction x cos theta)]"
looks like a good start. Here F = 150, friction = μmg. The result of the calc is the total force on the mass, which you can use in F = ma to find the acceleration. Once you have that, you can use accelerated motion formulas to find the Vf and then the Ek to complete part (a).
 


ok i got part a which was

533.0427

how do i get part B
i tried

V= square root 2(553.0427)/11
but got the wrong answer
 


Didn't you find the Vf in part (a) ? Once you found the acceleration, you would have used d = Vi*t + .5*a*t² to find the time and then
Vf = Vi + a*t to get Vf. Only after knowing Vf are you in a position to do the energy calc for the (a) answer.
 

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