- #1

- 1,462

- 44

## Homework Statement

A tension force of 2.50 N acts horizontally on a 2.00 kg block. The block accelerates at 0.750 m/s

^{2}. What is the force of kinetic friction?

## Homework Equations

F

_{net}= ma = F

_{1}+ F

_{2}+ F

_{3}+

**. . .**

## The Attempt at a Solution

I define the reference frame to be, in terms of an x axis, positive to the right and negative to the left. Thus, the force of tension is positive because I define it to be in the right direction, and the force of friction should be negative since it opposes the tension.

##F_{fx}## = force of friction

##F_{tx}## = tension

##F_{netx} = ma = F_{tx} + F_{fx}##

##F_{fx} = ma - F_{tx}##

##\left \| \vec{F_{f}} \right \|cos(180^{\circ}) = ma - \left \| \vec{F_{t}} \right \|cos(0^{\circ})##

##-\left \| \vec{F_{f}} \right \| = ma - \left \| \vec{F_{t}} \right \|cos(0^{\circ})##

##\left \| \vec{F_{f}} \right \| = \left \| \vec{F_{t}} \right \|cos(0^{\circ}) - ma##

##\left \| \vec{F_{f}} \right \| = (2.50 ~N) - (2.00 ~kg)(0.750 ~m/s^{2})##

##\left \| \vec{F_{f}} \right \| = (2.50 ~N) - (1.50~N)##

##\left \| \vec{F_{f}} \right \| = 1~N##

As we can see, I get positive 1 N rather than -1 N. Why is this? What am I doing wrong?