Work, energy and power Worked Solutions — Further Maths Maths

Fully worked, step-by-step solutions to Further Maths Work, energy and power questions. See exactly how to solve problems on work-energy-power, work-done-by-a-force, work-against-friction, work-energy-principle.

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Further Maths70 questionsStep-by-step solutions
Question 1
2 markseasy
A crate of mass 55 kg is modelled as a particle. The crate moves in a straight line on a horizontal floor from a point AA to a point BB, where AB=8AB=8 m. The resistance to motion is constant and has magnitude 1212 N. A constant force of magnitude 1717 N acts on the crate in the direction of motion. The crate is at rest at AA and passes through BB with speed 44 m s−1^{-1}. Take g=9.8 m s−2g=9.8\ \text{m s}^{-2}. Find the work done by the constant force as the crate moves from AA to BB.

Worked solution

  1. Write down the work-energy principle for the motion from AA to BB

    Wdrive−Wresistance=ΔKE+ΔPEW_{\text{drive}}-W_{\text{resistance}}=\Delta\text{KE}+\Delta\text{PE}

    Work supplied minus work lost to resistances equals the gain in kinetic plus potential energy.

  2. Find the work done by the applied force

    Wdrive=Fd=17×8=136 JW_{\text{drive}}=Fd=17\times 8=136\ \text{J}

    The force acts along the direction of motion, so cos⁡θ=1\cos\theta=1.

  3. State the final answer

    W=136 JW=136\ \text{J}

    This is the required quantity, correctly signed and with its units.

Answer
W=136 JW=136\ \text{J}
Question 2
2 markseasy
A box of mass 66 kg is modelled as a particle. The box moves in a straight line on a horizontal surface from a point AA to a point BB, where AB=15AB=15 m. The surface is rough and the coefficient of friction between the box and the surface is 0.50.5. A constant force of magnitude 34.434.4 N acts on the box in the direction of motion. The box is at rest at AA and passes through BB with speed 55 m s−1^{-1}. Take g=9.8 m s−2g=9.8\ \text{m s}^{-2}. Find the work done by the constant force as the box moves from AA to BB.

Worked solution

  1. Write down the work-energy principle for the motion from AA to BB

    Wdrive−Wresistance=ΔKE+ΔPEW_{\text{drive}}-W_{\text{resistance}}=\Delta\text{KE}+\Delta\text{PE}

    Work supplied minus work lost to resistances equals the gain in kinetic plus potential energy.

  2. Find the work done by the applied force

    Wdrive=Fd=34.4×15=516 JW_{\text{drive}}=Fd=34.4\times 15=516\ \text{J}

    The force acts along the direction of motion, so cos⁡θ=1\cos\theta=1.

  3. Note that there is no change in height

    ΔPE=0\Delta\text{PE}=0

    The motion is horizontal, so no work is done against gravity.

  4. State the final answer

    W=516 JW=516\ \text{J}

    This is the required quantity, correctly signed and with its units.

Answer
W=516 JW=516\ \text{J}
Question 3
2 markseasy
A block of mass 88 kg is modelled as a particle. The block moves in a straight line on a horizontal surface from a point AA to a point BB, where AB=25AB=25 m. The resistance to motion is constant and has magnitude 55 N. A constant force of magnitude 2121 N acts on the block in the direction of motion. The block is at rest at AA and passes through BB with speed 1010 m s−1^{-1}. Take g=9.8 m s−2g=9.8\ \text{m s}^{-2}. Find the work done against the resistance to motion as the block moves from AA to BB.

Worked solution

  1. Write down the work-energy principle for the motion from AA to BB

    Wdrive−Wresistance=ΔKE+ΔPEW_{\text{drive}}-W_{\text{resistance}}=\Delta\text{KE}+\Delta\text{PE}

    Work supplied minus work lost to resistances equals the gain in kinetic plus potential energy.

  2. Find the work done against the constant resistance

    Wresistance=Rd=5×25=125 JW_{\text{resistance}}=Rd=5\times 25=125\ \text{J}

    A constant resistance RR acting over a distance dd removes RdRd joules.

  3. State the final answer

    Wresistance=125 JW_{\text{resistance}}=125\ \text{J}

    This is the required quantity, correctly signed and with its units.

Answer
Wresistance=125 JW_{\text{resistance}}=125\ \text{J}
Question 4
2 markseasy
A parcel of mass 1010 kg is modelled as a particle. The parcel moves in a straight line on a horizontal floor from a point AA to a point BB, where AB=6AB=6 m. The floor is rough and the coefficient of friction between the parcel and the floor is 0.250.25. A constant force of magnitude 3232 N acts on the parcel in the direction of motion. The parcel is at rest at AA and passes through BB with speed 33 m s−1^{-1}. Take g=9.8 m s−2g=9.8\ \text{m s}^{-2}. Find the work done against friction as the parcel moves from AA to BB.

Worked solution

  1. Write down the work-energy principle for the motion from AA to BB

    Wdrive−Wresistance=ΔKE+ΔPEW_{\text{drive}}-W_{\text{resistance}}=\Delta\text{KE}+\Delta\text{PE}

    Work supplied minus work lost to resistances equals the gain in kinetic plus potential energy.

  2. Find the work done against friction

    Wfriction=μmgd=0.25×10×9.8×6=147 JW_{\text{friction}}=\mu mgd=0.25\times 10\times 9.8\times 6=147\ \text{J}

    This energy is lost as heat, so it is SUBTRACTED in the work-energy principle, never added.

  3. Note that there is no change in height

    ΔPE=0\Delta\text{PE}=0

    The motion is horizontal, so no work is done against gravity.

  4. State the final answer

    Wfriction=147 JW_{\text{friction}}=147\ \text{J}

    This is the required quantity, correctly signed and with its units.

Answer
Wfriction=147 JW_{\text{friction}}=147\ \text{J}
Question 5
2 markseasy
A sledge of mass 1212 kg is modelled as a particle. The sledge moves in a straight line on a horizontal surface from a point AA to a point BB, where AB=12AB=12 m. The surface is smooth. A constant force of magnitude 5050 N acts on the sledge in the direction of motion. The sledge is at rest at AA and passes through BB with speed 1010 m s−1^{-1}. Take g=9.8 m s−2g=9.8\ \text{m s}^{-2}. Find the increase in the kinetic energy of the sledge as it moves from AA to BB.

Worked solution

  1. Write down the work-energy principle for the motion from AA to BB

    Wdrive−Wresistance=ΔKE+ΔPEW_{\text{drive}}-W_{\text{resistance}}=\Delta\text{KE}+\Delta\text{PE}

    Work supplied minus work lost to resistances equals the gain in kinetic plus potential energy.

  2. Find the change in kinetic energy

    ΔKE=12mv2−12mu2=600−0=600 J\Delta\text{KE}=\tfrac{1}{2}mv^{2}-\tfrac{1}{2}mu^{2}=600-0=600\ \text{J}

    A negative value would mean the sledge has slowed down.

  3. State the final answer

    ∣ΔKE∣=600 J\left|\Delta\text{KE}\right|=600\ \text{J}

    This is the required quantity, correctly signed and with its units.

Answer
∣ΔKE∣=600 J\left|\Delta\text{KE}\right|=600\ \text{J}

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