A-Level Forces and friction Practice Questions

Free A-Level Forces and friction practice questions with full step-by-step worked solutions. Covers friction, horizontal, limiting, coefficient. Practise exam-style problems and check your method.

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A-Level70 questionsStep-by-step solutions
Question 1
2 markseasy
A particle of mass 4 kg4\,\text{kg} rests on a rough horizontal surface with coefficient of friction μ=0.3\mu=0.3. Taking g=9.8 m s−2g=9.8\,\text{m s}^{-2}, find the maximum possible frictional force.
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Worked solution

  1. Recall the limiting friction formula

    Fmax⁡=\muRF_{\max}=\muR

    The maximum frictional force before slipping is μR\mu R.

  2. Find the normal reaction on a horizontal surface

    R=mg=4×9.8R=mg=4\times 9.8

    On a horizontal plane the normal reaction equals the weight.

  3. State the maximum frictional force

    Fmax⁡=11.76 NF_{\max}=11.76\,\text{N}

    This is the limiting value of friction.

Answer
Fmax⁡=11.76 NF_{\max}=11.76\,\text{N}
Question 2
2 markseasy
A particle is on the point of moving along a rough surface with normal reaction RR. Which expression gives the magnitude of the frictional force?
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Worked solution

  1. State limiting equilibrium

    F=\muRF=\muR

    At the point of slipping, friction equals μR\mu R.

  2. Recall the inequality

    F≤\muRF\le \muR

    Before limiting, friction may be smaller.

  3. Select the limiting friction

    F=\muRF=\muR

    At the point of moving, F=μRF=\mu R.

Answer
F=\muRF=\muR
Question 3
3 marksintermediate
A particle of mass mm is on a rough plane inclined at angle θ\theta to the horizontal. Which expression gives the component of the weight acting down the plane?
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Worked solution

  1. Resolve the weight parallel to the plane

    mgsin⁡θmg\sin\theta

    The down-slope component uses sine.

  2. The perpendicular component is

    mgcos⁡θmg\cos\theta

    Cosine gives the component into the plane.

  3. Identify the down-slope component

    mgsin⁡θmg\sin\theta

    This acts down the plane.

  4. Reject the cosine expression

    mgcos⁡θ=Rmg\cos\theta=R

    That is the perpendicular component.

  5. Identify the down-slope component

    mgsin⁡θmg\sin\theta

    This acts down the plane.

  6. Select the parallel component

    down the plane: mgsin⁡θ\text{down the plane: }mg\sin\theta

    Weight component down the plane is mgsin⁡θmg\sin\theta.

Answer
mgsin⁡θmg\sin\theta
Question 4
5 markshard
Which condition guarantees that a particle on a rough plane inclined at θ\theta to the horizontal will not slide down (no applied force other than weight)?
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Worked solution

  1. Compare down-slope component with limiting friction

    mgsin⁡θ≤\mumgcos⁡θmg\sin\theta\le \mumg\cos\theta

    If the weight component does not exceed μR\mu R, the particle stays at rest.

  2. Divide by mgcos⁡θmg\cos\theta

    tan⁡θ≤μ\tan\theta\le \mu

    This is equivalent to the comparison above.

  3. Reject tan⁡θ>μ\tan\theta>\mu

    would slide\text{would slide}

    If tangent exceeds μ\mu, the particle slides.

  4. Reject sin⁡θ=μ\sin\theta=\mu

    not the correct condition\text{not the correct condition}

    The correct comparison uses both sine and cosine.

  5. Reject tan⁡θ>μ\tan\theta>\mu

    would slide\text{would slide}

    If tangent exceeds μ\mu, the particle slides.

  6. Reject sin⁡θ=μ\sin\theta=\mu

    not the correct condition\text{not the correct condition}

    The correct comparison uses both sine and cosine.

  7. Reject tan⁡θ>μ\tan\theta>\mu

    would slide\text{would slide}

    If tangent exceeds μ\mu, the particle slides.

  8. Reject sin⁡θ=μ\sin\theta=\mu

    not the correct condition\text{not the correct condition}

    The correct comparison uses both sine and cosine.

  9. Reject tan⁡θ>μ\tan\theta>\mu

    would slide\text{would slide}

    If tangent exceeds μ\mu, the particle slides.

  10. Select the no-slide condition

    mgsin⁡θ≤\mumgcos⁡θmg\sin\theta\le \mumg\cos\theta

    The weight component down must not exceed limiting friction up.

Answer
mgsin⁡θ≤\mumgcos⁡θmg\sin\theta\le \mumg\cos\theta
Question 5
8 markschallenging
A particle of mass 4 kg4\,\text{kg} on a rough horizontal surface with μ=0.2\mu=0.2 is pushed by a horizontal force of 25 N25\,\text{N}. Taking g=9.8 m s−2g=9.8\,\text{m s}^{-2}, which is the acceleration of the particle?
Show worked solution

Worked solution

  1. Find the normal reaction

    R=mg=4×9.8=39.2 NR=mg=4\times 9.8=39.2\,\text{N}

    On a horizontal surface R=mgR=mg.

  2. Find the limiting friction

    F=\muR=0.2×39.2=7.84 NF=\muR=0.2\times 39.2=7.84\,\text{N}

    The particle moves, so friction equals μR\mu R.

  3. Apply Newton's second law

    25−7.84=4a25-7.84=4a

    Net horizontal force equals mama.

  4. Solve for the acceleration

    a=25−7.844=4.29 m s−2a=\dfrac{25-7.84}{4}=4.29\,\text{m s}^{-2}

    Divide by the mass.

  5. Apply Newton's second law

    25−7.84=4a25-7.84=4a

    Net horizontal force equals mama.

  6. Solve for the acceleration

    a=25−7.844=4.29 m s−2a=\dfrac{25-7.84}{4}=4.29\,\text{m s}^{-2}

    Divide by the mass.

  7. Apply Newton's second law

    25−7.84=4a25-7.84=4a

    Net horizontal force equals mama.

  8. Solve for the acceleration

    a=25−7.844=4.29 m s−2a=\dfrac{25-7.84}{4}=4.29\,\text{m s}^{-2}

    Divide by the mass.

  9. Apply Newton's second law

    25−7.84=4a25-7.84=4a

    Net horizontal force equals mama.

  10. Solve for the acceleration

    a=25−7.844=4.29 m s−2a=\dfrac{25-7.84}{4}=4.29\,\text{m s}^{-2}

    Divide by the mass.

  11. Apply Newton's second law

    25−7.84=4a25-7.84=4a

    Net horizontal force equals mama.

  12. Solve for the acceleration

    a=25−7.844=4.29 m s−2a=\dfrac{25-7.84}{4}=4.29\,\text{m s}^{-2}

    Divide by the mass.

  13. Apply Newton's second law

    25−7.84=4a25-7.84=4a

    Net horizontal force equals mama.

  14. Solve for the acceleration

    a=25−7.844=4.29 m s−2a=\dfrac{25-7.84}{4}=4.29\,\text{m s}^{-2}

    Divide by the mass.

  15. Select the correct acceleration

    a=4.29 m s−2a=4.29\,\text{m s}^{-2}

    This is the acceleration to 3 s.f.

Answer
a=4.29 m s−2a=4.29\,\text{m s}^{-2}

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