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Newton's Laws of Motion

Applying Newton's laws to simple situations

  • 3 things to know
  • Unit AS 1

Overview

Recall questions on the general method for a mechanics problem, worked examples on a person on scales in an accelerating lift, a trolley pushing a second trolley, a car towing a trailer, a block on a smooth slope, a rocket at lift-off, and the development of terminal velocity.

What you need to know

Answer each one before you reveal it.

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Key term

Outline the method used to apply Newton's laws to a mechanics problem.

Numbers

A 1000 kg car tows a 500 kg trailer with a driving force of 4500 N and negligible resistance. Calculate the acceleration.

Key term

Explain the development of terminal velocity for a falling object using Newton's laws.

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    Outline the method used to apply Newton's laws to a mechanics problem.

    Model answer

    Choose the single body to be considered, identify every force acting on that body, choose one direction as positive, add the forces along that direction to find the resultant, and then apply F = ma to find the unknown quantity.

  2. 2

    A person of mass 70 kg stands on scales in a lift accelerating upwards at 1.5 m s⁻². Calculate the contact force from the floor, taking g as 9.81 m s⁻².

    Model answer

    The resultant must be upwards, so N − mg = ma giving N = 70 × (9.81 + 1.5) = 792 N; checking back, (792 − 687) ÷ 70 = 1.50 m s⁻², which is the acceleration given.

  3. 3

    A person of mass 70 kg stands on scales in a lift accelerating downwards at 1.5 m s⁻². Calculate the contact force from the floor, taking g as 9.81 m s⁻².

    Model answer

    The resultant must be downwards, so mg − N = ma giving N = 70 × (9.81 − 1.5) = 582 N; checking back, (687 − 582) ÷ 70 = 1.50 m s⁻², which is the acceleration given.

  4. 4

    A person of mass 70 kg stands on scales in a lift travelling upwards at constant velocity. Calculate the contact force from the floor, taking g as 9.81 m s⁻².

    Model answer

    Constant velocity means zero resultant force, so the contact force equals the weight, N = 70 × 9.81 = 687 N; checking back, 687 − 687 = 0 N, so there is indeed no acceleration.

  5. 5

    Explain why a person feels heavier during the moment a lift starts to move upwards.

    Model answer

    The sensation of weight comes from the contact force of the floor, and while the lift accelerates upwards that force must exceed the person's weight in order to produce the upward resultant force, so it feels greater than usual.

  6. 6

    A 10 N horizontal force is applied to a 3.0 kg trolley which is in contact with, and pushes, a 2.0 kg trolley on a frictionless surface. Calculate their common acceleration.

    Model answer

    The two move together, so a = F ÷ (m₁ + m₂) = 10 ÷ 5.0 = 2.0 m s⁻² in the direction of the applied force; checking back, 5.0 × 2.0 = 10 N, which is the applied force given.

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