A-Level Applications of forces Practice Questions

Free A-Level Applications of forces practice questions with full step-by-step worked solutions. Covers inclined-plane, resolving, normal-reaction, acceleration. Practise exam-style problems and check your method.

inclined-planeresolvingnormal-reactionaccelerationsmoothcomponents
A-Level70 questionsStep-by-step solutions
Question 1
2 markseasy
A particle of mass 2 kg2\,\text{kg} is on a smooth plane inclined at 30∘30^\circ to the horizontal. Taking g=9.8 m s−2g=9.8\,\text{m s}^{-2}, find the component of the weight acting down the plane, to 3 significant figures.
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Worked solution

  1. Resolve the weight parallel to the plane

    mgsin⁡θmg\sin\theta

    The component of weight acting down the smooth plane.

  2. Substitute the values

    mgsin⁡θ=2×9.8×sin⁡30∘mg\sin\theta=2\times 9.8\times\sin 30^\circ

    Insert the mass, gg and the angle.

  3. State the component down the plane

    mgsin⁡θ=9.8 Nmg\sin\theta=9.8\,\text{N}

    This is the parallel component to 3 s.f.

Answer
mgsin⁡θ=9.8 Nmg\sin\theta=9.8\,\text{N}
Question 2
2 markseasy
Two particles are connected by a light inextensible string over a smooth pulley. Which statement about the tension is correct while the system moves?
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Worked solution

  1. The string is light and inextensible

    T same throughoutT\ \text{same throughout}

    A light string has the same tension at every point.

  2. The pulley is smooth

    no friction at pulley\text{no friction at pulley}

    The tension is the same on both sides.

  3. Select the correct statement

    T is the same throughout the stringT\ \text{is the same throughout the string}

    One tension value throughout a light string.

Answer
T is the same throughoutT\ \text{is the same throughout}
Question 3
3 marksintermediate
For a particle on a smooth inclined plane, how does the normal reaction RR compare with the weight mgmg?
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Worked solution

  1. Resolve perpendicular to the plane

    R=mgcos⁡θR=mg\cos\theta

    Only the perpendicular component of weight is balanced.

  2. For an inclined plane

    0<cos⁡θ<10<\cos\theta<1

    The angle is between 0∘0^\circ and 90∘90^\circ.

  3. Compare with the weight

    R<mgR<mg

    Multiplying by cos⁡θ\cos\theta reduces the value.

  4. Reject equality unless horizontal

    R=mg⇔θ=0R=mg\Leftrightarrow\theta=0

    On a slope the normal reaction is less than the weight.

  5. Compare with the weight

    R<mgR<mg

    Multiplying by cos⁡θ\cos\theta reduces the value.

  6. Select the correct comparison

    R<mgR<mg

    The normal reaction is less than the weight on an incline.

Answer
R<mgR<mg
Question 4
5 markshard
A particle is pulled up a smooth plane by a force PP parallel to the plane. Which equation correctly applies Newton's second law up the plane?
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Worked solution

  1. Take up the plane as positive

    P up, mgsin⁡θ downP\ \text{up},\ mg\sin\theta\ \text{down}

    The applied force acts up; the weight component acts down.

  2. No friction on a smooth plane

    F=0F=0

    There is no frictional force.

  3. Apply F=maF=ma up the plane

    P−mgsin⁡θ=maP-mg\sin\theta=ma

    Net force up the plane equals mama.

  4. Reject P+mgsin⁡θ=maP+mg\sin\theta=ma

    mgsin⁡θ opposes motion upmg\sin\theta\ \text{opposes motion up}

    The weight component acts down the plane.

  5. Apply F=maF=ma up the plane

    P−mgsin⁡θ=maP-mg\sin\theta=ma

    Net force up the plane equals mama.

  6. Reject P+mgsin⁡θ=maP+mg\sin\theta=ma

    mgsin⁡θ opposes motion upmg\sin\theta\ \text{opposes motion up}

    The weight component acts down the plane.

  7. Apply F=maF=ma up the plane

    P−mgsin⁡θ=maP-mg\sin\theta=ma

    Net force up the plane equals mama.

  8. Reject P+mgsin⁡θ=maP+mg\sin\theta=ma

    mgsin⁡θ opposes motion upmg\sin\theta\ \text{opposes motion up}

    The weight component acts down the plane.

  9. Apply F=maF=ma up the plane

    P−mgsin⁡θ=maP-mg\sin\theta=ma

    Net force up the plane equals mama.

  10. Select the correct equation

    P−mgsin⁡θ=maP-mg\sin\theta=ma

    Newton's second law up a smooth plane.

Answer
P−mgsin⁡θ=maP-mg\sin\theta=ma
Question 5
8 markschallenging
A particle slides down a smooth plane inclined at 45∘45^\circ to the horizontal. Taking g=9.8 m s−2g=9.8\,\text{m s}^{-2}, what is the acceleration down the plane?
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Worked solution

  1. Apply Newton's second law down the plane

    mgsin⁡θ=mamg\sin\theta=ma

    Weight component down equals mama.

  2. Cancel the mass

    a=gsin⁡θa=g\sin\theta

    The mass cancels.

  3. Substitute θ=45∘\theta=45^\circ

    a=9.8sin⁡45∘a=9.8\sin 45^\circ

    Insert gg and the angle.

  4. Evaluate

    a=6.93 m s−2a=6.93\,\text{m s}^{-2}

    The acceleration to 3 s.f.

  5. Substitute θ=45∘\theta=45^\circ

    a=9.8sin⁡45∘a=9.8\sin 45^\circ

    Insert gg and the angle.

  6. Evaluate

    a=6.93 m s−2a=6.93\,\text{m s}^{-2}

    The acceleration to 3 s.f.

  7. Substitute θ=45∘\theta=45^\circ

    a=9.8sin⁡45∘a=9.8\sin 45^\circ

    Insert gg and the angle.

  8. Evaluate

    a=6.93 m s−2a=6.93\,\text{m s}^{-2}

    The acceleration to 3 s.f.

  9. Substitute θ=45∘\theta=45^\circ

    a=9.8sin⁡45∘a=9.8\sin 45^\circ

    Insert gg and the angle.

  10. Evaluate

    a=6.93 m s−2a=6.93\,\text{m s}^{-2}

    The acceleration to 3 s.f.

  11. Substitute θ=45∘\theta=45^\circ

    a=9.8sin⁡45∘a=9.8\sin 45^\circ

    Insert gg and the angle.

  12. Evaluate

    a=6.93 m s−2a=6.93\,\text{m s}^{-2}

    The acceleration to 3 s.f.

  13. Substitute θ=45∘\theta=45^\circ

    a=9.8sin⁡45∘a=9.8\sin 45^\circ

    Insert gg and the angle.

  14. Evaluate

    a=6.93 m s−2a=6.93\,\text{m s}^{-2}

    The acceleration to 3 s.f.

  15. Select the correct acceleration

    a=6.93 m s−2a=6.93\,\text{m s}^{-2}

    Acceleration down a smooth 45∘45^\circ plane.

Answer
a=6.93 m s−2a=6.93\,\text{m s}^{-2}

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