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Zorluk: OrtaNewton's Third Law (Action/Reaction)

Two warehouse workers are sitting on identical, frictionless wheeled carts facing each other. Worker X and his cart have a combined mass of 120 kg120 \text{ kg}, while Worker Y and his cart have a combined mass of 60 kg60 \text{ kg}. Worker X pushes against Worker Y's cart with a steady force of 240 N240 \text{ N}. Which of the following best describes the forces and resulting accelerations experienced by both workers during the push?

  1. Both workers experience an equal force of 240 N240 \text{ N} in opposite directions, but Worker Y accelerates twice as fast as Worker X.Cevap
  2. B
    Worker Y experiences a 240 N240 \text{ N} force, while Worker X experiences a 120 N120 \text{ N} reaction force due to the difference in their masses.
  3. C
    Worker X remains stationary during the push because his significantly greater inertia resists any backward reaction force.
  4. D
    Both workers experience an equal force of 240 N240 \text{ N}, but Worker X accelerates twice as fast because greater mass leads to greater acceleration.

Cevap

Both workers experience an equal force of 240 N240 \text{ N} in opposite directions, but Worker Y accelerates twice as fast as Worker X.
The correct answer accurately combines Newton's Third and Second Laws. By the Third Law, the action force of 240 N240 \text{ N} exerted by Worker X creates an equal and opposite reaction force of 240 N240 \text{ N} on Worker X. By the Second Law (F=maF = ma), acceleration is inversely proportional to mass. Since Worker Y has half the mass (60 kg60 \text{ kg}) of Worker X (120 kg120 \text{ kg}), the same 240 N240 \text{ N} force will cause Worker Y to accelerate exactly twice as fast.

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1
Determine the force acting on both workers using Newton's Third Law.
Both Worker X and Worker Y experience a 240 N240 \text{ N} force in opposite directions.
Newton's Third Law states that for every action, there is an equal and opposite reaction. The push of 240 N240 \text{ N} generates a reaction force of exactly 240 N240 \text{ N} on the person pushing.
2
Calculate the acceleration of Worker X using Newton's Second Law (a=Fma = \frac{F}{m}).
Acceleration of Worker X = 240 N120 kg=2 m/s2\frac{240 \text{ N}}{120 \text{ kg}} = 2 \text{ m/s}^2.
Acceleration depends on the net force applied and the total mass of the object.
3
Calculate the acceleration of Worker Y using Newton's Second Law (a=Fma = \frac{F}{m}).
Acceleration of Worker Y = 240 N60 kg=4 m/s2\frac{240 \text{ N}}{60 \text{ kg}} = 4 \text{ m/s}^2.
Worker Y has the same magnitude of force applied but half the mass, resulting in a different acceleration.
4
Compare the accelerations of the two workers.
Worker Y (4 m/s24 \text{ m/s}^2) accelerates exactly twice as fast as Worker X (2 m/s22 \text{ m/s}^2).
The inversely proportional relationship between mass and acceleration means the object with half the mass achieves twice the acceleration under an identical force.

Anahtar Kavram

Interaction of Newton's Third Law (Force Pairs) and Second Law (F=maF = ma)
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Konular

Newton's Third Law (Action/Reaction)

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