A 1.5-kg cart is pulled with a force of 7.3 N at an angle of 40 degrees above the horizontal. If a kinetic friction force of 3.2 N acts against the motion, the cart's acceleration along the horizontal surface will be?

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A 1.5-kg cart is pulled with a force of 7.3 N at an angle of 40 degrees above the horizontal. If a kinetic friction force of 3.2 N acts against the motion, the cart's acceleration along the horizontal surface will be?

Physics
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This is a simple balance of forces.\[\sum F = m*a = F - F_f\]
what is the F=? and Ff=?
i don't get it..

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Okay. From Newton's first law, The net force equals the mass times the acceleration. F is the applied force, F_f is the frictional force. Since frictional force always opposes the motion, it must be directed in the opposite direction as the applied force.
Newton's second law rather.
yes. Oh.. so it's Fa - Ff = m*a = F
what's the angel's role?
It is just\[m*a = F_a - F_f\]
ohh.
Ah. I misread the problem. We need to find the component of the applied force that is directed along the path of motion. \[F_p = F_a \sin(\theta)\]\[m*a = F_p - F_f\]
what is Fp?
I told you what F_p is. \[F_p = F_a \sin(\theta)\]It is the component of the force that acts along the path of motion.
oh but like is it like applied force..
Fp = 7.3 N sin(40 = 5.44 m*a = Fp - Ff 1.5*a = 5.44 - 3.2 1.5*a = 2.24 a = 1.49?
Yes.
Do you understand why?
no, I don't. Plus it's not in the answer provided.. please explain D: I'm so bad at physics i don't even know why i took it >_<"
First, begin every problem like this with a free body diagram. |dw:1327287228367:dw|Now, we need to break up the applied force into its normal components. That is, x and y based on the drawn coordinate system (a coordinate system should be included in every free body diagram)\[F_x = F_a \cos(\theta)\]\[F_y = F_a \sin(\theta)\](Whoops! I gave you the wrong expression for the applied force. I'm sorry.) We are interested in the horizontal acceleration. Or the acceleration in the x-direction. Now, let's write a balance of forces. \[m*a_x = F_a \cos(\theta) - F_f\]
ohh.. I'm kind of getting it.. thank you!!
ahh.. m*ax = Facos(theta)-Ff 1.5 * ax = 7.3cos(40)-(-3.2) ax = -1.67/1.5 ax = -1.113?
1.5 * a_x = 7.3 cos(40) - 3.2 Watch you signs. From the above free body diagram, we can see that F_f opposes the applied force and the resulting motion. Therefore, when setting up the force balance equation, we already account for it being negative.

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