(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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.

2. Relevant equations

F_{f}= muF_{n}

Vf^{2}= Vi^{2}+ 2ad

3. The attempt at a solution

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

therefore : Fapp_{x}= 359 cos 43 = 262.56N

Fn = mg , therefore

F_{f}= 0.15(80)(9.8) = 117.6N

sumF_{x}= 262.56 - 117.6 = 144.96N

which means:::: ma = 144.96N

80a = 144.96

a = 1.812

given Vf^{2}= Vi^{2}+ 2ad, and vi = 0 (rest)

then Vf^{2}= 2 (1.812)(1.5)

Vf = 2.33 m/s

-------> This is what Ive done on the question, im having doubts though, can anyone please verify / correct anywhere Im wrong? Please and Thank you

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# Homework Help: Weight pulled on Incline

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