A 8 kg block is moving with an initial speed of 6 m/s on a rough horizontal surface. If the force is 12 N, approximately how far does the block travel before it stops?

Question

A 8 kg block is moving with an initial speed of 6 m/s on a rough horizontal surface. If the force is 12 N, approximately how far does the block travel before it stops?
*I have: V^2=Vi^2+2ax
         0=(6 m/s)^2 + 2(-9.8 m/s^2)(x)
        -36 m^2/s^2 = -19.6 m/s^2(x)
         x = 1.83 m.
*Is this correct? Thank you in advance for your time. JL

Henry · Answer

g is used when the motion is vertical.
The block is moving horizontally. Therefore, the acceleration must be calculated:

Fk = Ma, a = Fk/M = -12/8 = -1.5 m/s^2.

V^2 = Vo^2 + 2a*d.
a = -1.5 m/s^2.

Fk is negative, because it opposed the
motion.