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Difficulty: MediumMechanics and Motion

A robotic transport carrier with a mass of 200 kg200\text{ kg} starts from rest and accelerates uniformly along a straight track, reaching a final velocity of 15 m/s15\text{ m/s} in 5 s5\text{ s}. What is the magnitude of the net force acting on the carrier?

  1. A
    60 N60\text{ N}
  2. 600 N600\text{ N}Answer
  3. C
    6000 N6{}000\text{ N}
  4. D
    300 N300\text{ N}

Answer

600 N600\text{ N}
The correct answer is 600 N600\text{ N}. First, find the constant acceleration by dividing the change in velocity by elapsed time: a=15 m/s5 s=3 m/s2a = \frac{15\text{ m/s}}{5\text{ s}} = 3\text{ m/s}^2. Next, use Newton's second law of motion (F=maF = ma) to find the net force: F=200 kg×3 m/s2=600 NF = 200\text{ kg} \times 3\text{ m/s}^2 = 600\text{ N}.

Step-by-Step Solution

1
Calculate the linear acceleration using the kinematic formula a=ΔvΔta = \frac{\Delta v}{\Delta t}.
a=15 m/s−0 m/s5 s=3 m/s2a = \frac{15\text{ m/s} - 0\text{ m/s}}{5\text{ s}} = 3\text{ m/s}^2
Acceleration represents the rate at which velocity changes over time.
2
Apply Newton's second law of motion (Fnet=m×aF_{\text{net}} = m \times a).
Fnet=200 kg×3 m/s2=600 NF_{\text{net}} = 200\text{ kg} \times 3\text{ m/s}^2 = 600\text{ N}
The net force required to accelerate an object is equal to the product of its mass and its acceleration.

Key Concept

Newton's Second Law of Motion and Kinematic Acceleration
Estimated Time:45s
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Topics

Mechanics and Motion

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