Calculating Work and Speed on a Frictionless Incline

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SUMMARY

The discussion focuses on calculating work and speed for an 18 kg block on a 32° frictionless incline under the influence of a horizontal force of 150 N. The work done by the gravitational force is determined using the formula W = mgsin(θ) multiplied by the distance of 2.1 m. The normal force does no work, resulting in a value of zero for part (c). Finally, the speed of the block after the displacement is calculated based on the net work done on the block.

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Homework Statement


How much work is done by the horizontal force Fp = 150 N on the 18 kg block of the figure when the force pushes the block 2.1 m up along the 32° frictionless incline? (b) How much work is done by the gravitational force on the block during this displacement? (c) How much work is done by the normal force? (d) What is the speed of the block (assume that it is zero initially) after this displacement? m/s

Homework Equations


[Hint: Work-energy involves net work done.]http://www.webassign.net/gianpse4/7-29.gif

The Attempt at a Solution


I know the answer to (c) is zero and I know how to do the problem when the force is parallel to the incline, but I am not sure how the values change with the force being at an angle.
 
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For a its just Force x distance

For b you have to calculate the force of gravity which is mgsin[tex]\theta[/tex] then you multiply by the distance

And C its zero
 
Thanks, I figured it out ;)
 

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