anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
This is a simple balance of forces.\[\sum F = m*a = F - F_f\]
anonymous
  • anonymous
what is the F=? and Ff=?
anonymous
  • anonymous
i don't get it..

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
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.
anonymous
  • anonymous
Newton's second law rather.
anonymous
  • anonymous
yes. Oh.. so it's Fa - Ff = m*a = F
anonymous
  • anonymous
what's the angel's role?
anonymous
  • anonymous
It is just\[m*a = F_a - F_f\]
anonymous
  • anonymous
ohh.
anonymous
  • anonymous
Ah. I misread the problem. We need to find the component of the applied force that is directed along the path of motion. \[F_p = F_a \sin(\theta)\]\[m*a = F_p - F_f\]
anonymous
  • anonymous
what is Fp?
anonymous
  • anonymous
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.
anonymous
  • anonymous
oh but like is it like applied force..
anonymous
  • anonymous
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?
anonymous
  • anonymous
Yes.
anonymous
  • anonymous
Do you understand why?
anonymous
  • anonymous
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 >_<"
anonymous
  • anonymous
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. \[m*a_x = F_a \cos(\theta) - F_f\]
anonymous
  • anonymous
ohh.. I'm kind of getting it.. thank you!!
anonymous
  • anonymous
ahh.. m*ax = Facos(theta)-Ff 1.5 * ax = 7.3cos(40)-(-3.2) ax = -1.67/1.5 ax = -1.113?
anonymous
  • anonymous
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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.