Asked by Jose Luis
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
*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
Answers
Answered by
Henry
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.
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.
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