Calculate Horizontal Force at an Incline

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SUMMARY

The discussion focuses on calculating the horizontal force exerted on a 3.54 kg block being pulled by a cord with a force of 11.90 N at an angle of 16.0 degrees above the horizontal. The gravitational force acting on the block is calculated as 34.7 N. To find the horizontal component of the force, it is essential to resolve the force into its components, specifically calculating Ftx, which represents the horizontal force acting on the block. The acceleration of the block is determined to be 3.36 m/s² based on the net force divided by mass.

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[SOLVED] Force at an Incline

A 3.54 kg block located on a horizontal frictionless floor is pulled by a cord that exerts a force F=11.90 N at an angle theta=16.0o above the horizontal, as shown. What is the speed of the block 6.10 seconds after it starts moving?

Fgravity = (3.54*9.8) = 34.7N

F/m = a ... 11.90/3.54 = 3.36 m/s^2

How would I calculate the horizontal force?
 
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First i think that the way you solved for your acceleration was wrong because if it is at an angle, you will need to find the components and set that to Fnet. For example, since the cord is exerting a force at 11.90 N at 16.0o that will be your Ft. you will need to find Ftx after that.
 

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