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anonymous
 4 years ago
A 1.5kg 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?
anonymous
 4 years ago
A 1.5kg 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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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0This is a simple balance of forces.\[\sum F = m*a = F  F_f\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0what is the F=? and Ff=?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Okay. 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Newton's second law rather.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes. Oh.. so it's Fa  Ff = m*a = F

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0what's the angel's role?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It is just\[m*a = F_a  F_f\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ah. I misread the problem. We need to find the component of the applied force that is directed along the path of motion. \[F_p = F_a \sin(\theta)\]\[m*a = F_p  F_f\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh but like is it like applied force..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Fp = 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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Do you understand why?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no, 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 >_<"

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0First, begin every problem like this with a free body diagram. dw:1327287228367:dwNow, 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 xdirection. Now, let's write a balance of forces. \[m*a_x = F_a \cos(\theta)  F_f\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ohh.. I'm kind of getting it.. thank you!!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ahh.. m*ax = Facos(theta)Ff 1.5 * ax = 7.3cos(40)(3.2) ax = 1.67/1.5 ax = 1.113?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.01.5 * 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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