F = m a
-12 = 8 a
a = -1.5 m/s^2
yes
0 = 6^2 + 2 a x
0 = 36 -3 x
x = 12 m
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
3 answers
Okay, I see now. Thanks Damon. I need to find acceleration first, then apply the kinematics formula. Thank you so Much!
You are welcome.
0 answers