Applications of forces Worked Solutions — A-Level Maths

Fully worked, step-by-step solutions to A-Level Applications of forces questions. See exactly how to solve problems on inclined-plane, resolving, normal-reaction, acceleration.

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.

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
A particle of mass 3 kg3\,\text{kg} is on a smooth plane inclined at 20∘20^\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.

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=3×9.8×sin⁡20∘mg\sin\theta=3\times 9.8\times\sin 20^\circ

    Insert the mass, gg and the angle.

  3. State the component down the plane

    mgsin⁡θ=10.1 Nmg\sin\theta=10.1\,\text{N}

    This is the parallel component to 3 s.f.

Answer
mgsin⁡θ=10.1 Nmg\sin\theta=10.1\,\text{N}
Question 3
2 markseasy
A particle of mass 2 kg2\,\text{kg} rests 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 magnitude of the normal reaction, to 3 significant figures.

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=2×9.8×cos⁡30∘R=2\times 9.8\times\cos 30^\circ

    Insert mm, gg and the angle.

  3. State the normal reaction

    R=17 NR=17\,\text{N}

    This is the normal reaction to 3 s.f.

Answer
R=17 NR=17\,\text{N}
Question 4
2 markseasy
A particle of mass 4 kg4\,\text{kg} rests 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 magnitude of the normal reaction, to 3 significant figures.

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=4×9.8×cos⁡45∘R=4\times 9.8\times\cos 45^\circ

    Insert mm, gg and the angle.

  3. State the normal reaction

    R=27.7 NR=27.7\,\text{N}

    This is the normal reaction to 3 s.f.

Answer
R=27.7 NR=27.7\,\text{N}
Question 5
2 markseasy
A particle of mass 2 kg2\,\text{kg} slides down 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 acceleration, to 3 significant figures.

Worked solution

  1. Apply Newton's second law down the plane

    mgsin⁡θ=mamg\sin\theta=ma

    On a smooth plane there is no friction; the weight component down the plane gives the acceleration.

  2. Cancel the mass

    a=gsin⁡θa=g\sin\theta

    The mass cancels from the equation.

  3. State the acceleration

    a=4.9 m s−2a=4.9\,\text{m s}^{-2}

    This is the acceleration down the smooth plane to 3 s.f.

Answer
a=4.9 m s−2a=4.9\,\text{m s}^{-2}

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