Further dynamics Worked Solutions — Further Maths Maths

Fully worked, step-by-step solutions to Further Maths Further dynamics questions. See exactly how to solve problems on further-dynamics, variable-force, work-done, a=v dv/dx.

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.

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 particle PP of mass 33 kg moves along the positive xx-axis. When the particle PP is at a distance xx metres from the origin OO, it is acted on by a resultant force of magnitude F=(8+2x)F=(8+2x) N directed in the direction of increasing xx. The particle PP passes through OO moving in the direction of increasing xx with speed 33 m s−1^{-1}. Find the work done by the force as the particle PP moves from OO to the point where x=6x=6.

Worked solution

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

    mvdvdx=8+2xmv\dfrac{dv}{dx}=8+2x

    The resultant force equals mass times acceleration.

  2. Separate the variables and integrate

    ∫3vmv dv=∫06(8+2x)dx\int_{3}^{v} mv\,dv=\int_{0}^{6}\left(8+2x\right)dx

    Collecting vv on the left and xx on the right makes both sides integrable.

  3. Work out the work done by the force

    ∫06(8+2x)dx=[x2+8x]06=84 J\int_{0}^{6}\left(8+2x\right)dx=\left[x^{2} + 8 x\right]_{0}^{6}=84\ \text{J}

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

  4. State the work done

    W=84 JW=84\ \text{J}

    The work done by the force equals this integral.

Answer
W=84 JW=84\ \text{J}
Question 3
2 markseasy
A small bead of mass 44 kg moves along the positive xx-axis. When the small bead is at a distance xx metres from the origin OO, it is acted on by a resultant force of magnitude F=(10+3x)F=(10+3x) N directed in the direction of increasing xx. The small bead passes through OO moving in the direction of increasing xx with speed 00 m s−1^{-1}. Find the acceleration of the small bead at the instant when x=4x=4.

Worked solution

  1. Model the motion and choose the form of the acceleration

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

    The force is a function of the displacement xx, so the useful form of the acceleration is v dv/dxv\,dv/dx.

  2. Write down Newton's second law for the small bead

    mvdvdx=10+3xmv\dfrac{dv}{dx}=10+3x

    The resultant force equals mass times acceleration.

  3. Evaluate the acceleration directly from F=maF=ma

    a=Fm=10+3x4a=\dfrac{F}{m}=\dfrac{10+3x}{4}

    Newton's second law gives the acceleration at any position from the force there.

  4. State the acceleration at the required point

    a=224=5.5 m s−2a=\dfrac{22}{4}=5.5\ \text{m s}^{-2}

    Substituting x=4x=4 gives the acceleration.

Answer
a=5.5 m s−2a=5.5\ \text{m s}^{-2}
Question 4
2 markseasy
A small block of mass 22 kg moves along the positive xx-axis. When the small block is at a distance xx metres from the origin OO, it is acted on by a resultant force of magnitude F=(5+4x)F=(5+4x) N directed in the direction of increasing xx. The small block passes through OO moving in the direction of increasing xx with speed 22 m s−1^{-1}. Find the acceleration of the small block at the instant when x=3x=3.

Worked solution

  1. Write down Newton's second law for the small block

    mvdvdx=5+4xmv\dfrac{dv}{dx}=5+4x

    The resultant force equals mass times acceleration.

  2. Evaluate the acceleration directly from F=maF=ma

    a=Fm=5+4x2a=\dfrac{F}{m}=\dfrac{5+4x}{2}

    Newton's second law gives the acceleration at any position from the force there.

  3. State the acceleration at the required point

    a=172=8.5 m s−2a=\dfrac{17}{2}=8.5\ \text{m s}^{-2}

    Substituting x=3x=3 gives the acceleration.

Answer
a=8.5 m s−2a=8.5\ \text{m s}^{-2}
Question 5
2 markseasy
A particle of mass 55 kg moves along a straight horizontal line. The only force acting on the particle is a resistance of magnitude 2v2v N, where vv m s−1^{-1} is the speed of the particle. The particle passes a point OO with speed 1212 m s−1^{-1}. Using a=vdvdxa=v\dfrac{dv}{dx}, find the speed of the particle when it has travelled 1010 m from OO.

Worked solution

  1. Write down the equation of motion

    mvdvdx=−2vm v\dfrac{dv}{dx}=-2v

    The resistance is the only force and it opposes the motion, hence the minus sign.

  2. Cancel a factor of vv and separate the variables

    m dv=−2 dxm\,dv=-2\,dx

    Dividing by vv leaves a constant deceleration in xx.

  3. Integrate and evaluate the speed

    v=12−25×10=8 m s−1v=12-\dfrac{2}{5}\times 10=8\ \text{m s}^{-1}

    Substituting the distance travelled gives the speed.

Answer
v=8 m s−1v=8\ \text{m s}^{-1}

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