(adsbygoogle = window.adsbygoogle || []).push({}); A 5.82 kg block located on a horizontal frictionless floor is pulled by a cord that exerts a force F = 12.80 N at an angle θ = 24.0° above the horizontal as shown. What is the speed of the block 5.90 seconds after it starts moving?

http://img239.imageshack.us/img239/1153/prob49dj7.gif [Broken]

F = ma

Fx = Fcosθ

Fy = Fsinθ

v(t) = v0 + at

Fx = 12.8cos24 = 11.69 N

Fx = m*ax ax = 2.01 m/s^2

vx(t) = 0 + 2.01(5.9) = 11.859 m/s

Fy = 12.8sin24 = 5.21 N

Fy = m*ay ay = 0.895 m/s^2

vy(t) = 0 + (.895)(5.9) = 5.2805 m/s

v(t)^2 = vx(t)^2 + vy(t)^2 = 11.859^2 + 5.2805^2

v(t) = 12.98 m/s

I entered 13.0 m/s into LONCAPA and it told me it was incorrect. What am I doing wrong?

Edit: Also tried a simpler way:

F=ma

12.8=5.82a

a=2.20m/s

v=v0+at

v=0+2.20(5.9)

v=12.98m/s

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# Homework Help: Motion and Forces

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