Calculating Work Done on a Box Being Pulled up a Ramp

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The problem involves calculating the work done on a box being pulled up a frictionless ramp with a tension of 350 N at a 50° angle to the horizontal, on a 20° incline, over a distance of 5 m. The formula used for work is W = Fd cos I. The initial calculations led to a work result of 1644 J, but a mistake was identified in the calculation where "250" should have been "350." Despite the typo, the final conclusion that the work done is 1644 J remains correct.
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Homework Statement


Suppose a box on a frictionless ramp is being pulled by a rope with a tension of 350 N making an angle of 50° to the horizontal ground. If the angle of incline of the ramp is 20°, and the box is pulled 5 m, determine the amount of work done. Hint: Draw a diagram and use vectors to represent the situation.


Homework Equations


W=FdcosI


The Attempt at a Solution


AC/AB= cos20degress
AC=ABcos20
=350cos20degrees
=39

W=HdcosI
= (5)(250cos20)
=1644J

The work done is 1644J

Is this solution correct?
 
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There is a mistake in your last calculation although the solution is correct. This should have been posted in Introductory physics, as I assume this was for a physics class?
 
Sorry I didn't know this was the wrong place to post. What is the mistake that I made?
 
A typo- you have "250" rather than "350".:biggrin:
 

Thread 'Family of lines that are at a distance of 5 from the origin'

I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
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