- #1
Mr Davis 97
- 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/s2. What is the force of kinetic friction?
Homework Equations
Fnet = ma = F1 + F2 + F3 + . . .
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?