A 3.0 kg object, moving with a constant velocity of 3.0 m/s, is acted upon by a force of 11 N in the direction of the motion for 4.0 s. What is the velocity (in meters/second) of the object at the end of this time?
Vf=Vi+a*t=Vi+Force/mass* time
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To find the velocity of the object at the end of the given time, we can use Newton's second law of motion. The formula is:
F = m * a
Where:
F = Force applied on the object
m = Mass of the object
a = Acceleration of the object
In this case, the force acting on the object is 11 N, and the mass of the object is 3.0 kg. Since the object is moving with a constant velocity, the net force acting on it is zero. Therefore, we can find the acceleration by rearranging the formula:
a = F/m
Substituting the given values, we get:
a = (11 N) / (3.0 kg)
a ≈ 3.67 m/s^2
Since the acceleration is constant, we can now use the equation for uniformly accelerated motion:
v = u + a * t
Where:
v = final velocity
u = initial velocity (in this case, 3.0 m/s)
a = acceleration (3.67 m/s^2)
t = time (4.0 s)
Substituting the values, we have:
v = (3.0 m/s) + (3.67 m/s^2) * (4.0 s)
v ≈ 3.0 m/s + 14.68 m/s
v ≈ 17.68 m/s
Thus, the velocity of the object at the end of 4.0 seconds is approximately 17.68 m/s.