Hard A-Level Applications of forces Questions

Challenging, exam-style A-Level Applications of forces questions with worked solutions. Stretch yourself on the hardest connected, inclined-plane, acceleration, tension problems.

connectedinclined-planeaccelerationtensionforceequilibrium
A-Level34 questionsStep-by-step solutions
Question 1
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?
Show worked solution

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}
Question 2
8 markschallenging
A particle is in equilibrium under three forces. Two have magnitudes 8 N8\,\text{N} at 50∘50^\circ and 6 N6\,\text{N} at 140∘140^\circ to the horizontal. Which is the magnitude of the third force?
Show worked solution

Worked solution

  1. State equilibrium

    F1+F2+F3=0\mathbf{F}_1+\mathbf{F}_2+\mathbf{F}_3=\mathbf{0}

    The forces sum to zero.

  2. Find the resultant of the two known forces

    R=F1+F2\mathbf{R}=\mathbf{F}_1+\mathbf{F}_2

    Add as vectors using components.

  3. The third force has magnitude ∣R∣|\mathbf{R}|

    F3=∣R∣F_3=|\mathbf{R}|

    It equals the magnitude of the resultant of the other two.

  4. Evaluate

    F3=10 NF_3=10\,\text{N}

    Use Pythagoras on the components.

  5. The third force has magnitude ∣R∣|\mathbf{R}|

    F3=∣R∣F_3=|\mathbf{R}|

    It equals the magnitude of the resultant of the other two.

  6. Evaluate

    F3=10 NF_3=10\,\text{N}

    Use Pythagoras on the components.

  7. The third force has magnitude ∣R∣|\mathbf{R}|

    F3=∣R∣F_3=|\mathbf{R}|

    It equals the magnitude of the resultant of the other two.

  8. Evaluate

    F3=10 NF_3=10\,\text{N}

    Use Pythagoras on the components.

  9. The third force has magnitude ∣R∣|\mathbf{R}|

    F3=∣R∣F_3=|\mathbf{R}|

    It equals the magnitude of the resultant of the other two.

  10. Evaluate

    F3=10 NF_3=10\,\text{N}

    Use Pythagoras on the components.

  11. The third force has magnitude ∣R∣|\mathbf{R}|

    F3=∣R∣F_3=|\mathbf{R}|

    It equals the magnitude of the resultant of the other two.

  12. Evaluate

    F3=10 NF_3=10\,\text{N}

    Use Pythagoras on the components.

  13. The third force has magnitude ∣R∣|\mathbf{R}|

    F3=∣R∣F_3=|\mathbf{R}|

    It equals the magnitude of the resultant of the other two.

  14. Evaluate

    F3=10 NF_3=10\,\text{N}

    Use Pythagoras on the components.

  15. Select the correct magnitude

    F3=10 NF_3=10\,\text{N}

    This is the magnitude to 3 s.f.

Answer
F3=10 NF_3=10\,\text{N}
Question 3
8 markschallenging
A particle of mass 5 kg5\,\text{kg} is on a smooth plane at 30∘30^\circ, connected to a hanging mass of 4 kg4\,\text{kg}. Taking g=9.8 m s−2g=9.8\,\text{m s}^{-2}, which is the acceleration of the system?
Show worked solution

Worked solution

  1. Compare driving forces

    4g=39.2 N, 5gsin⁡30∘=24.5 N4g=39.2\,\text{N},\ 5g\sin 30^\circ=24.5\,\text{N}

    Hanging weight versus slope component.

  2. Set up the system equation

    4g−5gsin⁡30∘=(5+4)a4g-5g\sin 30^\circ=(5+4)a

    Adding the equations of motion.

  3. Simplify the numerator

    4g−5gsin⁡30∘=14.7 N4g-5g\sin 30^\circ=14.7\,\text{N}

    Net driving effect.

  4. Divide by total mass

    a=14.79a=\dfrac{14.7}{9}

    Acceleration of the system.

  5. Simplify the numerator

    4g−5gsin⁡30∘=14.7 N4g-5g\sin 30^\circ=14.7\,\text{N}

    Net driving effect.

  6. Divide by total mass

    a=14.79a=\dfrac{14.7}{9}

    Acceleration of the system.

  7. Simplify the numerator

    4g−5gsin⁡30∘=14.7 N4g-5g\sin 30^\circ=14.7\,\text{N}

    Net driving effect.

  8. Divide by total mass

    a=14.79a=\dfrac{14.7}{9}

    Acceleration of the system.

  9. Simplify the numerator

    4g−5gsin⁡30∘=14.7 N4g-5g\sin 30^\circ=14.7\,\text{N}

    Net driving effect.

  10. Divide by total mass

    a=14.79a=\dfrac{14.7}{9}

    Acceleration of the system.

  11. Simplify the numerator

    4g−5gsin⁡30∘=14.7 N4g-5g\sin 30^\circ=14.7\,\text{N}

    Net driving effect.

  12. Divide by total mass

    a=14.79a=\dfrac{14.7}{9}

    Acceleration of the system.

  13. Simplify the numerator

    4g−5gsin⁡30∘=14.7 N4g-5g\sin 30^\circ=14.7\,\text{N}

    Net driving effect.

  14. Divide by total mass

    a=14.79a=\dfrac{14.7}{9}

    Acceleration of the system.

  15. Select the correct acceleration

    a=1.63 m s−2a=1.63\,\text{m s}^{-2}

    This is the acceleration to 3 s.f.

Answer
a=1.63 m s−2a=1.63\,\text{m s}^{-2}
Question 4
8 markschallenging
A particle of mass 8 kg8\,\text{kg} is on a smooth plane inclined at 45∘45^\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.
Show worked solution

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=8×9.8×sin⁡45∘mg\sin\theta=8\times 9.8\times\sin 45^\circ

    Insert the mass, gg and the angle.

  3. Note the direction

    down the plane\text{down the plane}

    This component acts down the slope.

  4. Write the weight

    W=mg=78.4 NW=mg=78.4\,\text{N}

    The full weight acts vertically.

  5. Confirm gg

    g=9.8 m s−2g=9.8\,\text{m s}^{-2}

    Gravitational field strength.

  6. Note the direction

    down the plane\text{down the plane}

    This component acts down the slope.

  7. Write the weight

    W=mg=78.4 NW=mg=78.4\,\text{N}

    The full weight acts vertically.

  8. Confirm gg

    g=9.8 m s−2g=9.8\,\text{m s}^{-2}

    Gravitational field strength.

  9. Note the direction

    down the plane\text{down the plane}

    This component acts down the slope.

  10. Write the weight

    W=mg=78.4 NW=mg=78.4\,\text{N}

    The full weight acts vertically.

  11. Confirm gg

    g=9.8 m s−2g=9.8\,\text{m s}^{-2}

    Gravitational field strength.

  12. Note the direction

    down the plane\text{down the plane}

    This component acts down the slope.

  13. Write the weight

    W=mg=78.4 NW=mg=78.4\,\text{N}

    The full weight acts vertically.

  14. Confirm gg

    g=9.8 m s−2g=9.8\,\text{m s}^{-2}

    Gravitational field strength.

  15. State the component down the plane

    mgsin⁡θ=55.4 Nmg\sin\theta=55.4\,\text{N}

    This is the parallel component to 3 s.f.

Answer
mgsin⁡θ=55.4 Nmg\sin\theta=55.4\,\text{N}
Question 5
8 markschallenging
A particle of mass 9 kg9\,\text{kg} rests on a smooth plane inclined at 60∘60^\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

    On a smooth plane the normal reaction balances the perpendicular component of weight.

  2. Substitute the values

    R=9×9.8×cos⁡60∘R=9\times 9.8\times\cos 60^\circ

    Insert mm, gg and the angle.

  3. Note the weight acts vertically

    W=mg=88.2 NW=mg=88.2\,\text{N}

    The weight is unchanged.

  4. State the direction of RR

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

    The normal reaction acts at right angles to the surface.

  5. Confirm gg

    g=9.8 m s−2g=9.8\,\text{m s}^{-2}

    Gravitational field strength.

  6. Note the weight acts vertically

    W=mg=88.2 NW=mg=88.2\,\text{N}

    The weight is unchanged.

  7. State the direction of RR

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

    The normal reaction acts at right angles to the surface.

  8. Confirm gg

    g=9.8 m s−2g=9.8\,\text{m s}^{-2}

    Gravitational field strength.

  9. Note the weight acts vertically

    W=mg=88.2 NW=mg=88.2\,\text{N}

    The weight is unchanged.

  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. Confirm gg

    g=9.8 m s−2g=9.8\,\text{m s}^{-2}

    Gravitational field strength.

  12. Note the weight acts vertically

    W=mg=88.2 NW=mg=88.2\,\text{N}

    The weight is unchanged.

  13. State the direction of RR

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

    The normal reaction acts at right angles to the surface.

  14. Confirm gg

    g=9.8 m s−2g=9.8\,\text{m s}^{-2}

    Gravitational field strength.

  15. State the normal reaction

    R=44.1 NR=44.1\,\text{N}

    This is the normal reaction to 3 s.f.

Answer
R=44.1 NR=44.1\,\text{N}

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