Question

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?

Answer 1

This is a simple balance of forces.\[\sum F = m*a = F - F_f\]

Answer 2

what is the F=? and Ff=?

Answer 3

i don't get it..

Answer 4

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.

Answer 5

Newton's second law rather.

Answer 6

yes. Oh.. so it's Fa - Ff = m*a = F

Answer 7

what's the angel's role?

Answer 8

It is just\[m*a = F_a - F_f\]

Answer 9

ohh.

Answer 10

Ah. I misread the problem. We need to find the component of the applied force that is directed along the path of motion. \[F_p = F_a \sin(\theta)\]\[m*a = F_p - F_f\]

Answer 11

what is Fp?

Answer 12

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.

Answer 13

oh but like is it like applied force..

Answer 14

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?

Answer 15

Yes.

Answer 16

Do you understand why?

Answer 17

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 >_<"

Answer 18

First, begin every problem like this with a free body diagram. 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. \[m*a_x = F_a \cos(\theta) - F_f\]