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
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.
Well, JL, your math seems to be on point! However, I believe there seems to be a bit of a numerical mishap. You see, when you squared the initial velocity of 6 m/s, it should be positive, not negative. So, let's try plugging that into the equation again.
V^2 = Vi^2 + 2ax
0 = (6 m/s)^2 + 2(-9.8 m/s^2)(x)
Simplifying this, we get:
0 = 36 m^2/s^2 - 19.6 m/s^2(x)
Now, since we want to find the distance x, we can solve for it:
19.6 m/s^2(x) = 36 m^2/s^2
x = 36 m^2/s^2 / 19.6 m/s^2
x ≈ 1.84 m
So there you have it! The block will travel approximately 1.84 meters before it comes to a stop. Keep those calculations rolling, JL!
Yes, your working and answer are correct. You used the equation V^2 = Vi^2 + 2ax, where V is the final velocity (which is 0 in this case since the block stops), Vi is the initial velocity (6 m/s), a is the acceleration (-9.8 m/s^2), and x is the distance traveled.
Substituting the given values into the equation:
0 = (6 m/s)^2 + 2(-9.8 m/s^2)(x)
0 = 36 m^2/s^2 - 19.6 m/s^2(x)
Simplifying the equation:
-36 m^2/s^2 = -19.6 m/s^2(x)
To solve for x, divide both sides of the equation by -19.6 m/s^2:
x = (-36 m^2/s^2) / (-19.6 m/s^2)
x ≈ 1.83 m
Therefore, the block travels approximately 1.83 meters before it stops.
Yes, your calculations are correct! You correctly applied the equation V^2 = Vi^2 + 2ax, where V is the final velocity (which is 0 since the block stops), Vi is the initial velocity (6 m/s), a is the acceleration (-9.8 m/s^2), and x is the distance traveled.
By substituting the given values into the equation, you correctly solved for x:
0 = (6 m/s)^2 + 2(-9.8 m/s^2)(x)
-36 m^2/s^2 = -19.6 m/s^2(x)
x = -36 m^2/s^2 / -19.6 m/s^2
x ≈ 1.83 m
Therefore, the block travels approximately 1.83 meters before it stops. Well done!