Velocity of a mass being pulled across a level surface

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SUMMARY

The discussion focuses on calculating the velocity of a crate being pulled across a level surface. A force of 172.0 N is applied at an angle of 28.0° while overcoming a frictional force of 124.0 N. The net force acting on the crate is determined to be 27.87 N in the horizontal direction. After applying the work-energy principle, the final velocity of the crate after being pulled 2.70 m is calculated to be 2.47 m/s.

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Homework Statement



A man pulls a crate of mass 67.0 kg across a level floor. He pulls with a force of 172.0 N at an angle of 28.0° above the horizontal. When the crate is moving, the frictional force between the floor and the crate has a magnitude of 124.0 N.
If the crate starts from rest, how fast will it be moving after the man has pulled it a distance of 2.70 m?

Homework Equations



W = Fsin(angle)
W = KE
KE = 1/2 m v^2


The Attempt at a Solution



I started by finding the net force so that I could use the work equation:

Fx = 172 cos(28) - 124 = 27.87 N
Fy = 172 sin(28) (Fn and W cancel) = 80.75 N
Fnet = sqrt(Fx^2 + fy^2) = 85.42 N

Work Equation

85.42(2.70)cos(28) = 203.6 J

KE Equation

203.6 J = .5 (67 kg) v^2

v= 2.47 m/s
 
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Careful, your frictional force acts in a opposing direction of your motion.. which is also angled.

And why does your [tex]F_n[/tex] and [tex]W[/tex] cancel? Weight comes straight down and your normal force is perpendicular to the surface.

Draw a free body diagram - it'll give you much better insight
 

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