Kinetic Energy & Work Question? (Halliday, Resnick, Walker, 7e, Ch. 7 #39)

AI Thread Summary
The discussion revolves around a physics problem involving a cart sliding along a frictionless rail, with a cord attached and a constant tension of 25.0N. The initial attempt at solving the problem involved calculating the angle of the cord, but it was pointed out that the angle changes as the cart moves. This oversight led to an incorrect calculation of work done, resulting in a discrepancy between the user's answer and the textbook's solution. After receiving guidance on considering the changing angle, the user was able to correct their approach. The conversation highlights the importance of accurately accounting for variable angles in physics problems.
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Homework Statement


A cord is attached to a cart that can slide along a frictionless horizontal rail aligned along an x axis. The left end of the cord is pulled over a pulley, or negligible mass and friction and at cord height h = 1.20m, so the cart slides from x1 = 3.00m to x2 = 1.00m. During the move, the tension in the cord is a constant 25.0N. what is the change in the kinetic energy of the cart during the move?


Homework Equations



W=Fxcos\Thetad ?
\DeltaK= 1/2mv22F - 1/2mv2I ?

The Attempt at a Solution



I attempted to find theta by doing tan-1(1.2/2) getting angle 31.96 degrees. I then plugged that into Fxcos(31.96)d = 25Ncos(31.96)2.00m = 42.9J which was my answer. But the book says it's 41.7J. I just can't figure it out...
 
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welcome to pf!

hi math_head7! welcome to pf! :smile:

(have a theta: θ and a delta: ∆ and a degree: ° :wink:)

i assume the lower end of the string stays 1.2 m below the pulley, so that the angle changes?

you've calculated it as if the angle stays the same :redface:

you'll need to integrate (alternatively, there is a trick using conservation) :wink:

try again! :smile:​
 
Thanks tiny-tim.

I didn't consider at all that θ was changing. Thanks to that advice I was able to figure it out.
 

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