A-Level Forces and Newton’s laws Practice Questions

Free A-Level Forces and Newton’s laws practice questions with full step-by-step worked solutions. Covers weight, forces, mass, newtons-second-law. Practise exam-style problems and check your method.

weightforcesmassnewtons-second-lawaccelerationforce
A-Level70 questionsStep-by-step solutions
Question 1
2 markseasy
A body has mass 5 kg5\,\text{kg}. Taking g=9.8 m s−2g=9.8\,\text{m s}^{-2}, find its weight.
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Worked solution

  1. Recall the connection between weight and mass

    W=mgW=mg

    Weight is the gravitational force acting on the mass.

  2. Substitute the mass and the value of g

    W=5×9.8W=5\times 9.8

    Insert the given mass and g.

  3. State the weight

    W=49 NW=49\,\text{N}

    This is the required weight.

Answer
W=49 NW=49\,\text{N}
Question 2
2 markseasy
A box rests on a horizontal floor. In which direction does the normal reaction from the floor act on the box?
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Worked solution

  1. Recall what the normal reaction is

    contact force perpendicular to the surface\text{contact force perpendicular to the surface}

    It acts at right angles to the surface.

  2. Identify the surface

    horizontal floor\text{horizontal floor}

    The floor is horizontal.

  3. State the direction

    vertically upward\text{vertically upward}

    The reaction is perpendicular to the floor, pushing up.

Answer
vertically upward\text{vertically upward}
Question 3
3 marksintermediate
Two forces F1=(7i−3j) N\mathbf{F}_1=(7\mathbf{i}-3\mathbf{j})\,\text{N} and F2=(−2i+8j) N\mathbf{F}_2=(-2\mathbf{i}+8\mathbf{j})\,\text{N} act on a particle. Find the resultant force as a vector.
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Worked solution

  1. Write the two forces in component form

    F1=7i−3j, F2=−2i+8j\mathbf{F}_1=7\mathbf{i}-3\mathbf{j},\ \mathbf{F}_2=-2\mathbf{i}+8\mathbf{j}

    Both forces are given in i, j components.

  2. Add the i-components

    Rx=7+(−2)=5R_x=7+(-2)=5

    Sum the horizontal components.

  3. Add the j-components

    Ry=−3+(8)=5R_y=-3+(8)=5

    Sum the vertical components.

  4. Combine into the resultant vector

    R=(5i+5j) N\mathbf{R}=(5\mathbf{i}+5\mathbf{j})\,\text{N}

    The resultant is the vector sum.

  5. Find the magnitude if required

    ∣R∣=52+52=7.07 N|\mathbf{R}|=\sqrt{5^{2}+5^{2}}=7.07\,\text{N}

    The magnitude follows from Pythagoras.

  6. State the resultant force

    R=(5i+5j) N\mathbf{R}=(5\mathbf{i}+5\mathbf{j})\,\text{N}

    This is the resultant as a vector.

Answer
R=(5i+5j) N\mathbf{R}=(5\mathbf{i}+5\mathbf{j})\,\text{N}
Question 4
5 markshard
A car accelerates forward because the road exerts a forward friction force on its driving wheels. What is the Newton's third law partner of this force?
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Worked solution

  1. State the given force

    road on tyres, forward\text{road on tyres, forward}

    This is the driving friction force.

  2. Recall Newton's third law

    equal and opposite, on the other body\text{equal and opposite, on the other body}

    The partner acts on the road, not the car.

  3. Identify the two bodies

    tyres and road\text{tyres and road}

    The interacting pair is the tyres and the road.

  4. Determine the direction

    backward\text{backward}

    The reaction is opposite to the given force.

  5. Write the partner force

    tyres on road, backward\text{tyres on road, backward}

    The tyres push back on the road.

  6. Check the equal-magnitude condition

    ∣F12∣=∣F21∣|F_{12}|=|F_{21}|

    The pair have equal magnitudes.

  7. Check they act on different bodies

    one on car, one on road\text{one on car, one on road}

    Third law pairs never act on the same body.

  8. Eliminate the air-resistance option

    that acts on the car\text{that acts on the car}

    Air resistance is a different, separate force.

  9. Eliminate the weight option

    weight is a gravity pair\text{weight is a gravity pair}

    The weight's partner is the car pulling the Earth.

  10. Select the correct partner

    friction of tyres on road, backward\text{friction of tyres on road, backward}

    This is the Newton's third law partner.

Answer
the backward friction of the tyres on the road\text{the backward friction of the tyres on the road}
Question 5
8 markschallenging
A particle is in equilibrium under three forces. Two are (7i−3j) N(7\mathbf{i}-3\mathbf{j})\,\text{N} and (−2i+5j) N(-2\mathbf{i}+5\mathbf{j})\,\text{N}. Which is the magnitude of the third force?
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Worked solution

  1. State the equilibrium condition

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

    The forces sum to zero.

  2. Add the i-components of the known forces

    Σx=7+(−2)=5\Sigma_x=7+(-2)=5

    Sum the horizontal components.

  3. Add the j-components of the known forces

    Σy=−3+5=2\Sigma_y=-3+5=2

    Sum the vertical components.

  4. Write the resultant of the known forces

    F1+F2=(5i+2j) N\mathbf{F}_1+\mathbf{F}_2=(5\mathbf{i}+2\mathbf{j})\,\text{N}

    Combine the two known forces.

  5. The third force cancels this

    F3=−(5i+2j)\mathbf{F}_3=-(5\mathbf{i}+2\mathbf{j})

    Equal and opposite to the resultant.

  6. Write its components

    F3=(−5i−2j) N\mathbf{F}_3=(-5\mathbf{i}-2\mathbf{j})\,\text{N}

    Negate each component.

  7. Write the magnitude formula

    ∣F3∣=(−5)2+(−2)2|\mathbf{F}_3|=\sqrt{(-5)^{2}+(-2)^{2}}

    Use Pythagoras.

  8. Simplify under the root

    ∣F3∣=29|\mathbf{F}_3|=\sqrt{29}

    Add the squares.

  9. Evaluate the magnitude

    ∣F3∣=5.39 N|\mathbf{F}_3|=5.39\,\text{N}

    The size of the third force.

  10. Verify the sum is zero

    (5−5)i+(2−2)j=0(5-5)\mathbf{i}+(2-2)\mathbf{j}=\mathbf{0}

    The three forces balance.

  11. Note the magnitude ignores sign

    ∣F3∣=∣−(5i+2j)∣|\mathbf{F}_3|=|{-}(5\mathbf{i}+2\mathbf{j})|

    Magnitude of a vector and its negative are equal.

  12. Eliminate the sum-of-magnitudes value

    ∣F3∣≠7.62|\mathbf{F}_3|\neq 7.62

    Do not add the magnitudes of the two forces.

  13. Eliminate the component values

    ∣F3∣≠5|\mathbf{F}_3|\neq 5

    5 is only the i-component.

  14. Check the arithmetic

    29=5.39\sqrt{29}=5.39

    Confirms the magnitude.

  15. Select the correct value

    ∣F3∣=5.39 N|\mathbf{F}_3|=5.39\,\text{N}

    This is the required magnitude to 3 s.f.

Answer
∣F3∣=5.39 N|\mathbf{F}_3|=5.39\,\text{N}

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