1/2 * 3 * v^2 < 600
v^2 < 400
|v| < 20
v^2 < 400
|v| < 20
Ek = 1/2mv^2
We are given that m = 3 kg and Ek < 600 J. Substituting these values into the equation, we get:
Ek = 1/2 * 3 kg * v^2
600 J = 1.5 kg * v^2
Now let's solve for v by isolating it on one side:
v^2 = 600 J / 1.5 kg
v^2 = 400 m^2/s^2
Taking the square root of both sides, we get:
v = ± √400 m/s
Therefore, the possible velocities for a 3 kg object with a kinetic energy of less than 600 J are ± 20 m/s.
Ek = 1/2 * m * v^2
Rearranging the formula:
v^2 = (2 * Ek) / m
v = √((2 * Ek) / m)
Now we can substitute the given values into the formula:
v = √((2 * 600 J) / 3 kg)
v = √(400 m^2/s^2)
v = ±20 m/s
The possible velocities for the 3 kg object, with a kinetic energy of less than 600 J, can be either +20 m/s or -20 m/s. The positive velocity represents the object moving in one direction, while the negative velocity represents the object moving in the opposite direction.