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A 63.9 kg is pulled by 125 N force at a 31.5-degree angle, while the water creates an 84.8 n force pulling directly backward. What is the y component of acceleration

 Jan 5, 2021
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I assume that the 31.5° is relative to the horizontal.

Then if the total force is F = 125N, then the force parallel to the water is

Fp = FCos31.5 = 125(0.85) = 106.625N

The resisting force by the water is 84.8N, so the net force is 106.625 – 84.8 = 21.825N

Then from F = ma

21.825 = 63.9a

So a = 0.342m/s^2 in the x-direction

Then the force normal to the water is

Fn = FSin31.5 = 125(0.5225) = 63.31N

Then again from F = ma

63.31N = 63.9a

So a = 63.31/63.9 = 0.991

So a = 0.991m/s^2 in the y-direction

 Jan 5, 2021

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