Final answer:
The magnitude of the 2.0-kg object's velocity at the end of the 3.0-second interval, after being acted upon by a constant force, is approximately 9.22 m/s, with the closest given option being 10.0 m/s. option d) 10.0 m/s is correct.
Step-by-step explanation:
A 2.0-kg object has an initial velocity of 4.0 m/s in the i direction at t=0. A constant resultant force of (2.0i + 4.0j) N acts on the object for 3.0 seconds. To find the magnitude of the object's velocity at the end of the 3.0-second interval, we must first determine the acceleration of the object due to the force.
The acceleration a can be found using Newton's second law, F = ma. Therefore, a = F/m = (2.0i + 4.0j) N / 2.0 kg = (1.0i + 2.0j) m/s2. The next step is to calculate the change in velocity, Δv = a × t = (1.0i + 2.0j) m/s2 × 3.0 s = (3.0i + 6.0j) m/s.
The initial velocity vector is 4.0i m/s. Adding the change in velocity to the initial velocity gives us the final velocity vector, v = 4.0i + 3.0i + 6.0j = 7.0i + 6.0j m/s. To calculate the magnitude of the final velocity, we use the Pythagorean theorem: |v| = √(7.02 + 6.02) m/s = √(49 + 36) m/s = √85 m/s ≈ 9.22 m/s. Therefore, the closest answer from the given options is 10.0 m/s (d).