Speed of Block 3.5s After Start Moving | Conical Pendulum

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SUMMARY

The discussion focuses on calculating the speed of a 3.46 kg block after 3.5 seconds of motion, influenced by a force of 13.9 N at an angle of 26.0° and a coefficient of kinetic friction of 0.09. The user applied the equations T*sin(theta) - mg = 0 for vertical forces and F + T*cos(theta) = ma for horizontal forces, resulting in an acceleration of 0.8790 m/s². The user expressed confusion regarding the acceleration value and its implications for the block's speed.

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  • Calculate the final speed of the block using the formula v = u + at
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A 3.46 kg block located on a horizontal floor is pulled by a cord that exerts a force F = 13.9 N at an angle theta = 26.0° above the horizontal, as shown in the Figure. The coefficient of kinetic friction between the block and the floor is 0.09. What is the speed of the block 3.5 s after it starts moving?

I'm having a bit of a problem with this. Mainly because I'm not sure how to answer the question. I got a decimal number less than 1 for the acceleration. Here are my two equations:

Y: T*sin(theta)-mg=0
X: F+T*cos(theta)=ma

a=.8790
 
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It's not centripetal motion, but I'm still having problems.
 

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