Hard A-Level Forces and friction Questions

Challenging, exam-style A-Level Forces and friction questions with worked solutions. Stretch yourself on the hardest inclined-plane, acceleration, force, limiting-equilibrium problems.

inclined-planeaccelerationforcelimiting-equilibriumfrictionhorizontal
A-Level34 questionsStep-by-step solutions
Question 1
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}
Question 2
8 markschallenging
A particle rests on a rough plane with μ=0.75\mu=0.75. Taking g=9.8 m s−2g=9.8\,\text{m s}^{-2}, at what angle of inclination to the horizontal is the particle on the point of sliding?
Show worked solution

Worked solution

  1. At limiting equilibrium

    mgsin⁡α=\mumgcos⁡αmg\sin\alpha=\mumg\cos\alpha

    Weight component down equals limiting friction.

  2. Rearrange

    tan⁡α=μ=0.75\tan\alpha=\mu=0.75

    Cancel mgmg and divide by cos⁡α\cos\alpha.

  3. Find the angle

    α=tan⁡−1(0.75)=36.9∘\alpha=\tan^{-1}(0.75)=36.9^\circ

    Use inverse tangent.

  4. Confirm gg cancels

    g not neededg\ \text{not needed}

    The limiting angle is independent of gg.

  5. Find the angle

    α=tan⁡−1(0.75)=36.9∘\alpha=\tan^{-1}(0.75)=36.9^\circ

    Use inverse tangent.

  6. Confirm gg cancels

    g not neededg\ \text{not needed}

    The limiting angle is independent of gg.

  7. Find the angle

    α=tan⁡−1(0.75)=36.9∘\alpha=\tan^{-1}(0.75)=36.9^\circ

    Use inverse tangent.

  8. Confirm gg cancels

    g not neededg\ \text{not needed}

    The limiting angle is independent of gg.

  9. Find the angle

    α=tan⁡−1(0.75)=36.9∘\alpha=\tan^{-1}(0.75)=36.9^\circ

    Use inverse tangent.

  10. Confirm gg cancels

    g not neededg\ \text{not needed}

    The limiting angle is independent of gg.

  11. Find the angle

    α=tan⁡−1(0.75)=36.9∘\alpha=\tan^{-1}(0.75)=36.9^\circ

    Use inverse tangent.

  12. Confirm gg cancels

    g not neededg\ \text{not needed}

    The limiting angle is independent of gg.

  13. Find the angle

    α=tan⁡−1(0.75)=36.9∘\alpha=\tan^{-1}(0.75)=36.9^\circ

    Use inverse tangent.

  14. Confirm gg cancels

    g not neededg\ \text{not needed}

    The limiting angle is independent of gg.

  15. Select the correct angle

    α=36.9∘\alpha=36.9^\circ

    This is the limiting angle to 3 s.f.

Answer
α=36.9∘\alpha=36.9^\circ
Question 3
8 markschallenging
A particle of mass 5 kg5\,\text{kg} rests on a rough plane inclined at 30∘30^\circ with μ=0.2\mu=0.2. Taking g=9.8 m s−2g=9.8\,\text{m s}^{-2}, which statement is correct?
Show worked solution

Worked solution

  1. Find the weight component down the plane

    mgsin⁡30∘=5×9.8×0.5=24.5 Nmg\sin 30^\circ=5\times 9.8\times 0.5=24.5\,\text{N}

    Resolve the weight parallel to the plane.

  2. Find the limiting friction up the plane

    Fmax⁡=\mumgcos⁡30∘=0.2×5×9.8×cos⁡30∘=8.49 NF_{\max}=\mumg\cos 30^\circ=0.2\times 5\times 9.8\times\cos 30^\circ=8.49\,\text{N}

    Maximum friction with R=mgcos⁡θR=mg\cos\theta.

  3. Compare the two values

    mgsin⁡θ>\mumgcos⁡θmg\sin\theta>\mumg\cos\theta

    This determines whether the particle slides.

  4. Conclude about motion

    the particle slides\text{the particle slides}

    Compare the parallel weight with limiting friction.

  5. Compare the two values

    mgsin⁡θ>\mumgcos⁡θmg\sin\theta>\mumg\cos\theta

    This determines whether the particle slides.

  6. Conclude about motion

    the particle slides\text{the particle slides}

    Compare the parallel weight with limiting friction.

  7. Compare the two values

    mgsin⁡θ>\mumgcos⁡θmg\sin\theta>\mumg\cos\theta

    This determines whether the particle slides.

  8. Conclude about motion

    the particle slides\text{the particle slides}

    Compare the parallel weight with limiting friction.

  9. Compare the two values

    mgsin⁡θ>\mumgcos⁡θmg\sin\theta>\mumg\cos\theta

    This determines whether the particle slides.

  10. Conclude about motion

    the particle slides\text{the particle slides}

    Compare the parallel weight with limiting friction.

  11. Compare the two values

    mgsin⁡θ>\mumgcos⁡θmg\sin\theta>\mumg\cos\theta

    This determines whether the particle slides.

  12. Conclude about motion

    the particle slides\text{the particle slides}

    Compare the parallel weight with limiting friction.

  13. Compare the two values

    mgsin⁡θ>\mumgcos⁡θmg\sin\theta>\mumg\cos\theta

    This determines whether the particle slides.

  14. Conclude about motion

    the particle slides\text{the particle slides}

    Compare the parallel weight with limiting friction.

  15. Select the correct statement

    the particle slides down the plane\text{the particle slides down the plane}

    Compare mgsin⁡θmg\sin\theta with μmgcos⁡θ\mu mg\cos\theta.

Answer
the particle slides\text{the particle slides}
Question 4
8 markschallenging
A particle of mass 8 kg8\,\text{kg} rests on a rough plane inclined at 55∘55^\circ to the horizontal. Taking g=9.8 m s−2g=9.8\,\text{m s}^{-2}, find the magnitude of the normal reaction, to 3 significant figures.
Show worked solution

Worked solution

  1. Resolve perpendicular to the plane

    R=mgcos⁡θR=mg\cos\theta

    The normal reaction balances the perpendicular component of the weight.

  2. Substitute the mass and angle

    R=8×9.8×cos⁡55∘R=8\times 9.8\times\cos 55^\circ

    Insert mm, gg and the angle to the horizontal.

  3. Note the weight acts vertically

    W=mg=78.4 NW=mg=78.4\,\text{N}

    The weight is unchanged by the slope.

  4. Identify the angle used

    θ=55∘\theta=55^\circ

    The plane is inclined at this angle.

  5. Confirm gg

    g=9.8 m s−2g=9.8\,\text{m s}^{-2}

    Gravitational field strength.

  6. State the direction of RR

    perpendicular to the plane\text{perpendicular to the plane}

    The normal reaction acts at right angles to the surface.

  7. Note the weight acts vertically

    W=mg=78.4 NW=mg=78.4\,\text{N}

    The weight is unchanged by the slope.

  8. Identify the angle used

    θ=55∘\theta=55^\circ

    The plane is inclined at this angle.

  9. Confirm gg

    g=9.8 m s−2g=9.8\,\text{m s}^{-2}

    Gravitational field strength.

  10. State the direction of RR

    perpendicular to the plane\text{perpendicular to the plane}

    The normal reaction acts at right angles to the surface.

  11. Note the weight acts vertically

    W=mg=78.4 NW=mg=78.4\,\text{N}

    The weight is unchanged by the slope.

  12. Identify the angle used

    θ=55∘\theta=55^\circ

    The plane is inclined at this angle.

  13. Confirm gg

    g=9.8 m s−2g=9.8\,\text{m s}^{-2}

    Gravitational field strength.

  14. State the direction of RR

    perpendicular to the plane\text{perpendicular to the plane}

    The normal reaction acts at right angles to the surface.

  15. State the normal reaction

    R=45 NR=45\,\text{N}

    This is the required normal reaction to 3 s.f.

Answer
R=45 NR=45\,\text{N}
Question 5
8 markschallenging
A particle of mass 10 kg10\,\text{kg} on a rough horizontal surface with μ=0.4\mu=0.4 is pushed by a horizontal force of 60 N60\,\text{N} and accelerates. Taking g=9.8 m s−2g=9.8\,\text{m s}^{-2}, find the acceleration, to 3 significant figures.
Show worked solution

Worked solution

  1. Find the normal reaction and limiting friction

    R=mg=10×9.8, Ff=\muR=39.2 NR=mg=10\times 9.8,\ F_f=\muR=39.2\,\text{N}

    On a horizontal surface R=mgR=mg; once sliding, friction equals μR\mu R.

  2. Apply Newton's second law horizontally

    60−Ff=10a60-F_f=10a

    The net horizontal force equals mass times acceleration.

  3. Substitute the known forces

    60−39.2=10a60-39.2=10a

    Insert the applied force and limiting friction.

  4. Rearrange for the acceleration

    a=60−39.210a=\dfrac{60-39.2}{10}

    Make aa the subject.

  5. Confirm F>\muRF>\muR

    60>39.260>39.2

    The applied force exceeds the limiting friction.

  6. Confirm gg

    g=9.8 m s−2g=9.8\,\text{m s}^{-2}

    Used in the normal reaction.

  7. Substitute the known forces

    60−39.2=10a60-39.2=10a

    Insert the applied force and limiting friction.

  8. Rearrange for the acceleration

    a=60−39.210a=\dfrac{60-39.2}{10}

    Make aa the subject.

  9. Confirm F>\muRF>\muR

    60>39.260>39.2

    The applied force exceeds the limiting friction.

  10. Confirm gg

    g=9.8 m s−2g=9.8\,\text{m s}^{-2}

    Used in the normal reaction.

  11. Substitute the known forces

    60−39.2=10a60-39.2=10a

    Insert the applied force and limiting friction.

  12. Rearrange for the acceleration

    a=60−39.210a=\dfrac{60-39.2}{10}

    Make aa the subject.

  13. Confirm F>\muRF>\muR

    60>39.260>39.2

    The applied force exceeds the limiting friction.

  14. Confirm gg

    g=9.8 m s−2g=9.8\,\text{m s}^{-2}

    Used in the normal reaction.

  15. State the acceleration

    a=2.08 m s−2a=2.08\,\text{m s}^{-2}

    This is the acceleration to 3 s.f.

Answer
a=2.08 m s−2a=2.08\,\text{m s}^{-2}

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