(adsbygoogle = window.adsbygoogle || []).push({}); Kinetic Friction Problem..Need Help!!

Okay...so I really think I'm doing this right..but the CAPA thing won't take my answer so there has to be something I'm doing wrong.

The Question is ...

A 32.0 kg box is being pulled along a horizontal surface by means of a rope that exerts a force of 265 N at an angle of 54.5o above the horizontal. The coefficient of kinetic friction between the box and the surface is 0.330. What is the acceleration of the box?

So I drew my free body diagram and I had the Normal Force and the Weight and cosine of the angle in the y direction and I had the kinetic friction (normal force times kinetic friction coefficient) and weight and sine of the angle in the X direction.

Then I worked it all out and I got

Fy = may

Fy = 0

N - W(cos54.5) = 0

N - mg(cos54.5) = 0

N = 32.0kg (9.81 m/s^2)(cos54.5)

N = 182.3

Fx = max

-fk + T - w(sin54.5) = max

T - (N x uk) - mg(sin54.5) = max

265 - (182.3 x 0.330) - (32kg)(9.81m/s^2)(sin54.5) = 32kg(ax)

then I solved It all and it worked out to be -1.59 m/s^2

but its saying its not right..so I'm confused beyond belief now.

so Please help me .. lol

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# Homework Help: Kinetic Friction ProblemNeed Help

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