What is the speed of the block after this displacement?

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SUMMARY

The discussion centers on calculating the speed of a 20 kg block after it is displaced 3.6 m up a 30° frictionless incline. The work done by the horizontal force is 467.7 J, the gravitational force does -352.8 J of work, and the normal force does 0 J of work. The total work done on the block is 114.9 J, which leads to the correct calculation of speed using the formula W = 0.5m(v^2). The correct speed of the block after the displacement is 6.77 m/s, not 6.83 m/s as initially calculated.

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meganw
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Homework Statement



I did the first 3 parts of the question, but I can't seem to get the right answer on the last one.

(a) How much work is done by the horizontal force Fp on the 20 kg block when the force pushes the block 3.6 m up the 30° frictionless incline?
467.7 J
(b) How much work is done by the gravitational force on the block during this displacement?
-352.8 J
(c) How much work is done by the normal force?
0 J
(d) What is the speed of the block (assume that it is zero initially) after this displacement?


Homework Equations


F=ma
K=.5m(v^2)

The Attempt at a Solution



I know that W = .5m(v^2) in this case, and since on D, (thats the one I have a question on), I'm trying to find Velocity, I can use this formula. I tried

467.7 = .5m(v^2)

and got v=6.83 but that's wrong.

Thank you so much for your help!

=)
 
Physics news on Phys.org
What's the TOTAL work done on the object?
 
Hey, I didn't think to do that, THANK YOU I got it right! =) =)
 

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