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 2kg2\,\text{kg} is on a smooth plane inclined at 3030^\circ to the horizontal. Taking g=9.8m s2g=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

    mgsinθmg\sin\theta

    The component of weight acting down the smooth plane.

  2. Substitute the values

    mgsinθ=2×9.8×sin30mg\sin\theta=2\times 9.8\times\sin 30^\circ

    Insert the mass, gg and the angle.

  3. State the component down the plane

    mgsinθ=9.8Nmg\sin\theta=9.8\,\text{N}

    This is the parallel component to 3 s.f.

Answer
mgsinθ=9.8Nmg\sin\theta=9.8\,\text{N}
Question 2
2 markseasy
A particle of mass 3kg3\,\text{kg} is on a smooth plane inclined at 2020^\circ to the horizontal. Taking g=9.8m s2g=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

    mgsinθmg\sin\theta

    The component of weight acting down the smooth plane.

  2. Substitute the values

    mgsinθ=3×9.8×sin20mg\sin\theta=3\times 9.8\times\sin 20^\circ

    Insert the mass, gg and the angle.

  3. State the component down the plane

    mgsinθ=10.1Nmg\sin\theta=10.1\,\text{N}

    This is the parallel component to 3 s.f.

Answer
mgsinθ=10.1Nmg\sin\theta=10.1\,\text{N}
Question 3
2 markseasy
A particle of mass 2kg2\,\text{kg} rests on a smooth plane inclined at 3030^\circ to the horizontal. Taking g=9.8m s2g=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=mgcosθR=mg\cos\theta

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

  2. Substitute the values

    R=2×9.8×cos30R=2\times 9.8\times\cos 30^\circ

    Insert mm, gg and the angle.

  3. State the normal reaction

    R=17NR=17\,\text{N}

    This is the normal reaction to 3 s.f.

Answer
R=17NR=17\,\text{N}
Question 4
2 markseasy
A particle of mass 4kg4\,\text{kg} rests on a smooth plane inclined at 4545^\circ to the horizontal. Taking g=9.8m s2g=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=mgcosθR=mg\cos\theta

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

  2. Substitute the values

    R=4×9.8×cos45R=4\times 9.8\times\cos 45^\circ

    Insert mm, gg and the angle.

  3. State the normal reaction

    R=27.7NR=27.7\,\text{N}

    This is the normal reaction to 3 s.f.

Answer
R=27.7NR=27.7\,\text{N}
Question 5
2 markseasy
A particle of mass 2kg2\,\text{kg} slides down a smooth plane inclined at 3030^\circ to the horizontal. Taking g=9.8m s2g=9.8\,\text{m s}^{-2}, find the acceleration, to 3 significant figures.

Worked solution

  1. Apply Newton's second law down the plane

    mgsinθ=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=gsinθa=g\sin\theta

    The mass cancels from the equation.

  3. State the acceleration

    a=4.9m s2a=4.9\,\text{m s}^{-2}

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

Answer
a=4.9m s2a=4.9\,\text{m s}^{-2}

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