Further Maths Further dynamics Practice Questions

Free Further Maths Further dynamics practice questions with full step-by-step worked solutions. Covers further-dynamics, variable-force, work-done, a=v dv/dx. Practise exam-style problems and check your method.

further-dynamicsvariable-forcework-donea=v dv/dxresistancea=dv/dt
Further Maths70 questionsStep-by-step solutions
Question 1
2 markseasy
A particle of mass 22 kg moves along the positive xx-axis. When the particle is at a distance xx metres from the origin OO, it is acted on by a resultant force of magnitude F=(6+3x)F=(6+3x) N directed in the direction of increasing xx. The particle passes through OO moving in the direction of increasing xx with speed 44 m s−1^{-1}. Find the work done by the force as the particle moves from OO to the point where x=5x=5.
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Worked solution

  1. Write down Newton's second law for the particle

    mvdvdx=6+3xmv\dfrac{dv}{dx}=6+3x

    The resultant force equals mass times acceleration.

  2. Work out the work done by the force

    ∫05(6+3x)dx=[3x22+6x]05=67.5 J\int_{0}^{5}\left(6+3x\right)dx=\left[\frac{3 x^{2}}{2} + 6 x\right]_{0}^{5}=67.5\ \text{J}

    This definite integral is the work done by the variable force.

  3. State the work done

    W=67.5 JW=67.5\ \text{J}

    The work done by the force equals this integral.

Answer
W=67.5 JW=67.5\ \text{J}
Question 2
2 markseasy
A bead of mass 44 kg moves in a straight line. At time tt seconds the resultant force acting on the bead has magnitude F=(5+8t)F=(5+8t) N and acts in the direction of motion. When t=0t=0 the bead has speed 44 m s−1^{-1}. Find the magnitude of the impulse of the force on the bead during the first 33 seconds of the motion. Select the correct value from the options given.
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Worked solution

  1. Choose the form of the acceleration for a time-dependent force

    a=dvdta=\dfrac{dv}{dt}

    The force is a function of tt, so use a=dv/dta=dv/dt.

  2. Write down Newton's second law

    mdvdt=5+8tm\dfrac{dv}{dt}=5+8t

    The resultant force equals mass times acceleration.

  3. Recall that the impulse is the integral of the force over time

    J=∫03(5+8t)dtJ=\int_{0}^{3}\left(5+8t\right)dt

    Impulse is the time integral of a variable force.

  4. Evaluate the impulse

    J=[4t2+5t]03=51 N sJ=\left[4 t^{2} + 5 t\right]_{0}^{3}=51\ \text{N s}

    This definite integral is the magnitude of the impulse.

Answer
J=51 N sJ=51\ \text{N s}
Question 3
4 marksintermediate
A small ring of mass 22 kg moves in a straight line. At time tt seconds the resultant force acting on the small ring has magnitude F=(5+4t)F=(5+4t) N and acts in the direction of motion. When t=0t=0 the small ring has speed 33 m s−1^{-1}. Using a=dvdta=\dfrac{dv}{dt}, find the speed of the small ring when t=5t=5. Select the correct value from the options given.
Show worked solution

Worked solution

  1. Choose the form of the acceleration for a time-dependent force

    a=dvdta=\dfrac{dv}{dt}

    The force is a function of tt, so use a=dv/dta=dv/dt.

  2. Write down Newton's second law

    mdvdt=5+4tm\dfrac{dv}{dt}=5+4t

    The resultant force equals mass times acceleration.

  3. Integrate to find the speed as a function of time

    v=3+12∫0t(5+4t)dtv=3+\dfrac{1}{2}\int_{0}^{t}\left(5+4t\right)dt

    Integrating the acceleration and using v=3v=3 at t=0t=0 gives v(t)v(t).

  4. Simplify the expression for the speed

    v=t2+5t2+3v=t^{2} + \frac{5 t}{2} + 3

    This is the speed of the small ring at time tt.

  5. Recall the two forms of the acceleration

    a=dvdt=vdvdxa=\dfrac{dv}{dt}=v\dfrac{dv}{dx}

    The form to use depends on whether the force is given in terms of time or of displacement.

  6. Recall Newton's second law

    F=maF=ma

    The resultant force equals the mass times the acceleration at every instant.

  7. Substitute the required time

    v=40.5 m s−1v=40.5\ \text{m s}^{-1}

    Putting t=5t=5 into v(t)v(t) gives the speed.

Answer
v=40.5 m s−1v=40.5\ \text{m s}^{-1}
Question 4
6 markshard
A train of mass 20002000 kg moves along a straight horizontal road. The engine of the train works at a constant rate of 1600016000 W. When the train is moving with speed vv m s−1^{-1} the total resistance to motion has magnitude 8v28v^{2} N. Find the maximum speed of the train. Give your answer to 3 significant figures. Select the correct value from the options given.
Show worked solution

Worked solution

  1. Relate the driving force to the power and the speed

    F=PvF=\dfrac{P}{v}

    The engine works at a constant rate, so the driving force is P/vP/v.

  2. Apply the condition for the maximum speed

    a=0 ⇒ Pv=8v2a=0\ \Rightarrow\ \dfrac{P}{v}=8v^{2}

    At the maximum speed the driving force just balances the resistance.

  3. Rearrange for the speed

    v3=P8=160008v^{3}=\dfrac{P}{8}=\dfrac{16000}{8}

    Multiplying up gives a power of vv equal to P/kP/k.

  4. Recall the two forms of the acceleration

    a=dvdt=vdvdxa=\dfrac{dv}{dt}=v\dfrac{dv}{dx}

    The form to use depends on whether the force is given in terms of time or of displacement.

  5. Recall Newton's second law

    F=maF=ma

    The resultant force equals the mass times the acceleration at every instant.

  6. Recall the work done by a variable force

    W=∫F dxW=\int F\,dx

    The work of a force that varies with position is the integral of the force with respect to displacement.

  7. Recall the work-energy principle

    W=12mv2−12mu2W=\tfrac{1}{2}mv^{2}-\tfrac{1}{2}mu^{2}

    The work done by the resultant force equals the change in kinetic energy.

  8. Recall the impulse of a variable force

    J=∫F dtJ=\int F\,dt

    The impulse of a time-varying force is the integral of the force with respect to time.

  9. Recall the impulse-momentum principle

    J=mv−muJ=mv-mu

    The impulse of the resultant force equals the change in momentum.

  10. Recall that power is the rate of working

    P=FvP=Fv

    For a driving force FF at speed vv the power developed is FvFv.

  11. Solve for the maximum speed

    v=1600083=12.6 m s−1v=\sqrt[3]{\dfrac{16000}{8}}=12.6\ \text{m s}^{-1}

    Taking the (3)(3)th root gives the maximum speed.

Answer
v=12.6 m s−1v=12.6\ \text{m s}^{-1}
Question 5
9 markschallenging
A van of mass 24002400 kg moves along a straight horizontal road. The engine of the van works at a constant rate of 2800028000 W. When the van is moving with speed vv m s−1^{-1} the total resistance to motion has magnitude 7v27v^{2} N. Find the maximum speed of the van. Give your answer to 3 significant figures. Select the correct value from the options given.
Show worked solution

Worked solution

  1. Relate the driving force to the power and the speed

    F=PvF=\dfrac{P}{v}

    The engine works at a constant rate, so the driving force is P/vP/v.

  2. Apply the condition for the maximum speed

    a=0 ⇒ Pv=7v2a=0\ \Rightarrow\ \dfrac{P}{v}=7v^{2}

    At the maximum speed the driving force just balances the resistance.

  3. Rearrange for the speed

    v3=P7=280007v^{3}=\dfrac{P}{7}=\dfrac{28000}{7}

    Multiplying up gives a power of vv equal to P/kP/k.

  4. Recall the two forms of the acceleration

    a=dvdt=vdvdxa=\dfrac{dv}{dt}=v\dfrac{dv}{dx}

    The form to use depends on whether the force is given in terms of time or of displacement.

  5. Recall Newton's second law

    F=maF=ma

    The resultant force equals the mass times the acceleration at every instant.

  6. Recall the work done by a variable force

    W=∫F dxW=\int F\,dx

    The work of a force that varies with position is the integral of the force with respect to displacement.

  7. Recall the work-energy principle

    W=12mv2−12mu2W=\tfrac{1}{2}mv^{2}-\tfrac{1}{2}mu^{2}

    The work done by the resultant force equals the change in kinetic energy.

  8. Recall the impulse of a variable force

    J=∫F dtJ=\int F\,dt

    The impulse of a time-varying force is the integral of the force with respect to time.

  9. Recall the impulse-momentum principle

    J=mv−muJ=mv-mu

    The impulse of the resultant force equals the change in momentum.

  10. Recall that power is the rate of working

    P=FvP=Fv

    For a driving force FF at speed vv the power developed is FvFv.

  11. Recall the condition for the maximum speed

    a=0a=0

    At the maximum speed the acceleration is zero, so the driving force balances the resistance.

  12. Separate the variables before integrating

    ∫dvf(v)=∫dx\int \frac{dv}{f(v)}=\int dx

    A separable first-order equation is solved by collecting the speed on one side.

  13. State the value of gg used throughout

    g=9.8 m s−2g=9.8\ \text{m s}^{-2}

    All numerical answers use this value of the acceleration due to gravity.

  14. Check the units of each term

    [F]=N,[W]=J[F]=\text{N},\quad [W]=\text{J}

    Forces are in newtons and work and energy in joules.

  15. Note that a resistance opposes the motion

    R=−kvnR=-kv^{n}

    A resistive force always acts opposite to the velocity, so it reduces the speed.

  16. Solve for the maximum speed

    v=2800073=15.9 m s−1v=\sqrt[3]{\dfrac{28000}{7}}=15.9\ \text{m s}^{-1}

    Taking the (3)(3)th root gives the maximum speed.

Answer
v=15.9 m s−1v=15.9\ \text{m s}^{-1}

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