How Fast Does a 80 kg Weight Move After Being Pulled 1.5 m on a Rough Plane?

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An 80 kg weight is pulled 1.5 m on a rough horizontal plane with a force of 359 N at a 43-degree angle, resulting in a friction coefficient of 0.15. The applied force in the x direction is calculated to be 262.56 N, while the frictional force is 117.6 N, leading to a net force of 144.96 N. This net force is used to determine the acceleration, which is found to be 1.812 m/s². Using the kinematic equation, the final velocity after moving 1.5 m from rest is calculated to be 2.33 m/s. The calculations appear to be correct, and the doubts expressed are addressed by confirming the accuracy of the steps taken.
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Homework Statement



A weight, mass m = 80 kg , is pulled over a rough horizontal plane, mu = 0.15, by a force of magnitude F = 359 N, making an angle of 43 degrees with the plane, away from the plane. Calculate the speed of the weight d= 1.5 m away from its initial position assuming that it started from rest.

Homework Equations



Ff = muFn

Vf2 = Vi2 + 2ad

The Attempt at a Solution



The applied force is evaluated in the x direction only (since the object remained on the floor)

therefore : Fappx = 359 cos 43 = 262.56N

Fn = mg , therefore
Ff = 0.15(80)(9.8) = 117.6N
sumFx = 262.56 - 117.6 = 144.96N

which means:::: ma = 144.96N
80a = 144.96
a = 1.812

given Vf2 = Vi2 + 2ad, and vi = 0 (rest)

then Vf2 = 2 (1.812)(1.5)

Vf = 2.33 m/s-------> This is what I've done on the question, I am having doubts though, can anyone please verify / correct anywhere I am wrong? Please and Thank you
 
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It seems right to me. Why do you have doubts?
 

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