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

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The discussion revolves around calculating the change in kinetic energy and final speed of a crate being pulled up a rough incline by a force. The initial attempts at solving the problem involved incorrect calculations and misunderstandings about the required parameters, such as the angle and coefficient of friction. A correct approach was suggested, emphasizing the need to determine the total force acting on the crate and subsequently using kinematic equations to find the final velocity. After resolving part A, the user sought clarification on how to proceed with part B, indicating confusion about the relationship between acceleration, time, and final velocity. The conversation highlights the importance of correctly applying physics principles to solve problems involving forces and motion.
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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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