Further Maths Work, energy and power Practice Questions

Free Further Maths Work, energy and power practice questions with full step-by-step worked solutions. Covers work-energy-power, work-done-by-a-force, work-against-friction, work-energy-principle. Practise exam-style problems and check your method.

work-energy-powerwork-done-by-a-forcework-against-frictionwork-energy-principleinclined-planepower
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.
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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 crate of mass 66 kg is modelled as a particle. The crate moves in a straight line on a horizontal surface from a point AA to a point BB, where AB=10AB=10 m. The surface is rough and the coefficient of friction between the crate and the surface is 0.250.25. A constant force of magnitude 17.417.4 N acts on the crate in the direction of motion. The crate 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 crate moves from AA to BB. Select the correct value from the options given.
Show worked solution

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=0.25×6×9.8×10=147 JW_{\text{friction}}=\mu mgd=0.25\times 6\times 9.8\times 10=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. Select the option that matches this value

    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 3
4 marksintermediate
A car of mass 14001400 kg is modelled as a particle. The car moves along a straight horizontal road. The resistance to motion is constant and has magnitude 900900 N. The engine of the car works at a constant rate of 68.2568.25 kW. The car is moving with speed 3535 m s−1^{-1} and has acceleration 0.750.75 m s−2^{-2}. Take g=9.8 m s−2g=9.8\ \text{m s}^{-2}. Find the magnitude of the driving force of the engine at this instant. Select the correct value from the options given.
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Worked solution

  1. Set up the model

    particle,g=9.8 m s−2\text{particle},\quad g=9.8\ \text{m s}^{-2}

    The car is modelled as a particle moving in a straight line.

  2. Recall the relationship between power, driving force and speed

    P=FvP=Fv

    The power developed by the engine is the driving force times the speed.

  3. Write down Newton's second law along the road

    F−R=maF-R=ma

    The resultant force along the direction of motion produces the acceleration.

  4. Convert the rate of working into watts

    68.25 kW=68250 W68.25\ \text{kW}=68250\ \text{W}

    Powers must be in watts before P=FvP=Fv is used with SI units.

  5. Use P=FvP=Fv to find the driving force at this instant

    F=Pv=6825035=1950 NF=\frac{P}{v}=\frac{68250}{35}=1950\ \text{N}

    The engine works at a constant rate, so the driving force falls as the speed rises.

  6. Check the rate of working against the energy budget

    P−Rv−mgvsin⁡α−mav=68250−31500−0−36750=0P-Rv-mgv\sin\alpha-mav=68250-31500-0-36750=0

    The engine supplies energy at exactly the rate at which it is used up by the resistance, by the climb and by the gain in kinetic energy.

  7. Select the option that matches this value

    F=1950 NF=1950\ \text{N}

    This is the required quantity, with the correct units.

Answer
F=1950 NF=1950\ \text{N}
Question 4
6 markshard
A motorcycle of mass 15001500 kg is modelled as a particle. The motorcycle moves up a straight road inclined at an angle α\alpha to the horizontal, where sin⁡α=114\sin\alpha=\frac{1}{14}. The resistance to motion is constant and has magnitude 600600 N. The engine of the motorcycle works at a constant rate of 43.243.2 kW. The motorcycle is moving with speed 1818 m s−1^{-1}. Take g=9.8 m s−2g=9.8\ \text{m s}^{-2}. Find the acceleration of the motorcycle at this instant. Select the correct value from the options given.
Show worked solution

Worked solution

  1. Set up the model

    particle,g=9.8 m s−2\text{particle},\quad g=9.8\ \text{m s}^{-2}

    The motorcycle is modelled as a particle moving in a straight line.

  2. Recall the relationship between power, driving force and speed

    P=FvP=Fv

    The power developed by the engine is the driving force times the speed.

  3. Write down Newton's second law along the road

    F−R−mgsin⁡α=maF-R-mg\sin\alpha=ma

    The resultant force along the direction of motion produces the acceleration.

  4. Find the component of the weight along the road

    mgsin⁡α=1500×9.8×114=1050 Nmg\sin\alpha=1500\times 9.8\times \frac{1}{14}=1050\ \text{N}

    It is sin⁡α\sin\alpha that resolves the weight ALONG the slope; cos⁡α\cos\alpha would resolve it perpendicular to the slope, which is not what is wanted here.

  5. Convert the rate of working into watts

    43.2 kW=43200 W43.2\ \text{kW}=43200\ \text{W}

    Powers must be in watts before P=FvP=Fv is used with SI units.

  6. Use P=FvP=Fv to find the driving force at this instant

    F=Pv=4320018=2400 NF=\frac{P}{v}=\frac{43200}{18}=2400\ \text{N}

    The engine works at a constant rate, so the driving force falls as the speed rises.

  7. Apply Newton's second law and solve for the acceleration

    a=F−R−mgsin⁡αm=7501500=0.5 m s−2a=\frac{F-R-mg\sin\alpha}{m}=\frac{750}{1500}=0.5\ \text{m s}^{-2}

    The resultant force divided by the mass gives the acceleration.

  8. Check the rate of working against the energy budget

    P−Rv−mgvsin⁡α−mav=43200−10800−18900−13500=0P-Rv-mgv\sin\alpha-mav=43200-10800-18900-13500=0

    The engine supplies energy at exactly the rate at which it is used up by the resistance, by the climb and by the gain in kinetic energy.

  9. Interpret the equation of motion physically

    Fdrive−R−mgsin⁡α=ma>0F_{\text{drive}}-R-mg\sin\alpha=ma>0

    The engine supplies more than the resistances require, and the surplus accelerates the motorcycle.

  10. State the units of the answer

    [a]=m s−2\left[a\right]=\text{m s}^{-2}

    Powers are in watts (or kW), forces in newtons, speeds in m s−1^{-1} and accelerations in m s−2^{-2}.

  11. Select the option that matches this value

    a=0.5 m s−2a=0.5\ \text{m s}^{-2}

    This is the required quantity, with the correct units.

Answer
a=0.5 m s−2a=0.5\ \text{m s}^{-2}
Question 5
9 markschallenging
A lorry of mass 12001200 kg is modelled as a particle. The lorry moves down a straight road inclined at an angle α\alpha to the horizontal, where sin⁡α=114\sin\alpha=\frac{1}{14}. The resistance to motion is constant and has magnitude 900900 N. The engine of the lorry works at a constant rate of 1.081.08 kW. The lorry is moving at its maximum speed. Take g=9.8 m s−2g=9.8\ \text{m s}^{-2}. Find the maximum speed of the lorry. Select the correct value from the options given.
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Worked solution

  1. Set up the model

    particle,g=9.8 m s−2\text{particle},\quad g=9.8\ \text{m s}^{-2}

    The lorry is modelled as a particle moving in a straight line.

  2. Recall the relationship between power, driving force and speed

    P=FvP=Fv

    The power developed by the engine is the driving force times the speed.

  3. Use the condition for maximum speed

    a=0  ⟹  Fresultant=0a=0\;\Longrightarrow\;F_{\text{resultant}}=0

    At maximum speed the lorry is no longer accelerating, so the resultant force along the road is zero: the driving force can only balance the resistances.

  4. Write down Newton's second law along the road

    F−R+mgsin⁡α=maF-R+mg\sin\alpha=ma

    The resultant force along the direction of motion produces the acceleration.

  5. Find the component of the weight along the road

    mgsin⁡α=1200×9.8×114=840 Nmg\sin\alpha=1200\times 9.8\times \frac{1}{14}=840\ \text{N}

    It is sin⁡α\sin\alpha that resolves the weight ALONG the slope; cos⁡α\cos\alpha would resolve it perpendicular to the slope, which is not what is wanted here.

  6. Convert the rate of working into watts

    1.08 kW=1080 W1.08\ \text{kW}=1080\ \text{W}

    Powers must be in watts before P=FvP=Fv is used with SI units.

  7. Evaluate the driving force from the equation of motion

    F=R−mgsin⁡α=60 NF=R-mg\sin\alpha=60\ \text{N}

    At this speed the engine must supply exactly this force.

  8. Use P=FvP=Fv to find the maximum speed

    v=PF=108060=18 m s−1v=\frac{P}{F}=\frac{1080}{60}=18\ \text{m s}^{-1}

    Dividing the rate of working by the driving force gives the speed.

  9. Check the rate of working against the energy budget

    P−Rv−mgvsin⁡α−mav=1080−16200+15120−0=0P-Rv-mgv\sin\alpha-mav=1080-16200+15120-0=0

    The engine supplies energy at exactly the rate at which it is used up by the resistance, by the climb and by the gain in kinetic energy.

  10. Interpret the maximum-speed condition physically

    Fdrive=R−mgsin⁡αF_{\text{drive}}=R-mg\sin\alpha

    Because the acceleration is zero at maximum speed there is no resultant force, so the driving force can do no more than balance the resistances; it is not free to be larger.

  11. State the units of the answer

    [vmax⁡]=m s−1\left[v_{\max}\right]=\text{m s}^{-1}

    Powers are in watts (or kW), forces in newtons, speeds in m s−1^{-1} and accelerations in m s−2^{-2}.

  12. Recall the definition of the work done by a constant force

    W=Fdcos⁡θW=Fd\cos\theta

    Only the component of the force along the displacement does any work.

  13. Recall the formula for kinetic energy

    KE=12mv2\text{KE}=\tfrac{1}{2}mv^{2}

    Kinetic energy is measured in joules when mm is in kg and vv in m s−1^{-1}.

  14. Recall the formula for gravitational potential energy

    PE=mgh\text{PE}=mgh

    Here hh is the vertical height gained, not the distance travelled.

  15. Recall the work-energy principle

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

    Work put in minus work lost to resistances equals the total gain in energy.

  16. Select the option that matches this value

    vmax⁡=18 m s−1v_{\max}=18\ \text{m s}^{-1}

    This is the required quantity, with the correct units.

Answer
vmax⁡=18 m s−1v_{\max}=18\ \text{m s}^{-1}

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