Mathematics · AQA 7357 · Topic R

Forces and Newton's laws

Forces and Newton's laws: resolving forces, inclined planes, connected particles, pulleys and friction.

  • 6 sub-topics
  • 1 in Year 12, 2 in Year 13
  • 78 practice questions
  • 6 animated lessons
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What the specification says

Every sub-topic AQA can examine in this topic, with its reference number from the official specification. The Y12 and Y13 tags show when each sub-topic is usually taught.

  1. R1Forces and Newton's first lawY12
  2. R2Newton's second law and resolving forcesY13
  3. R3Weight and motion in a straight line under gravityY13
  4. R4Newton's third law, pulleys and connected particlesY13
  5. R5Resultant forces and dynamics in a planeY13
  6. R6Friction, F ≤ μRY13

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Forces and Newton's laws

Easy

Newton's second law · Year 12 · 2 marks

  1. A car of mass 1000 kg has a driving force of 2500 N and resistance of 700 N. Find its acceleration.
    (2)
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  1. Resultant force = mass × acceleration.2500−700=1000a2500 - 700 = 1000a
  2. a=1.8a = 1.8 m s⁻²

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Forces and Newton's laws

Medium

Particle on an inclined plane · Year 12 · 4 marks

A particle of mass 2 kg is pulled up a smooth plane inclined at 30° to the horizontal by a force of 15 N acting parallel to the plane. Take g=9.8 m s−2g = 9.8\ \mathrm{m\,s^{-2}}.
  1. Find the acceleration of the particle.
    (4)
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  1. Resolve along the plane and use F=maF = ma. The weight's component down the slope is mgsin⁡30∘mg\sin 30^\circ.15−2gsin⁡30∘=2a15 - 2g\sin 30^\circ = 2a
  2. 2×9.8×0.5=9.82 \times 9.8 \times 0.5 = 9.8.15−9.8=2a15 - 9.8 = 2a
  3. The plane is smooth, so there is no friction to include.a=2.6 m s−2a = 2.6\ \mathrm{m\,s^{-2}} up the plane

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Forces and Newton's laws

Not needed for Year 12Hard

Pushing up a rough slope · Year 13 · 6 marks

  1. A 10 kg box is on a rough plane inclined at 20∘20^\circ (μ=0.4\mu = 0.4). Find the force parallel to the plane needed to make the box just about to move up.
    (6)
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  1. R=10gcos⁡20∘=92.1R = 10g\cos 20^\circ = 92.1 N
  2. Friction opposes the movement that is about to happen.Friction acts down the slope: F=0.4RF = 0.4R
  3. Forces up the slope balance forces down it.PP=10gsin⁡20∘{}= 10g\sin 20^\circ+0.4×10gcos⁡20∘{}+ 0.4 \times 10g\cos 20^\circ
  4. P=70.4P = 70.4 N

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