Calculate Horizontal Force at an Incline

[am08]

[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?

 

[MaZnFLiP]

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.

 

[Moetasim]

To calculate the horizontal force, we can use the formula Fhorizontal = Fapplied*cos(theta), where Fapplied is the applied force and theta is the angle at which it is applied. In this case, Fapplied is the given force of 11.90 N and theta is 16.0 degrees. Plugging these values into the formula, we get Fhorizontal = 11.90*cos(16.0) = 11.31 N. This is the horizontal component of the applied force, which is responsible for the horizontal acceleration of the block.