Hard Further Maths Work, energy and power Questions

Challenging, exam-style Further Maths Work, energy and power questions with worked solutions. Stretch yourself on the hardest work-energy-power, inclined-plane, work-energy-principle, work-against-friction problems.

work-energy-powerinclined-planework-energy-principlework-against-frictionpowerP=Fv
Further Maths34 questionsStep-by-step solutions
Question 1
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.
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 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}
Question 2
9 markschallenging
A block of mass 1212 kg is modelled as a particle. The block moves up a line of greatest slope of a plane inclined at an angle α\alpha to the horizontal, where sin⁡α=725\sin\alpha=\frac{7}{25}, from a point AA to a point BB, where AB=16AB=16 m. The plane is rough and the coefficient of friction between the block and the plane is 0.30.3. A constant force of magnitude 150150 N acts on the block in the direction of motion. The block is at rest at AA. Take g=9.8 m s−2g=9.8\ \text{m s}^{-2}. Find the speed of the block at BB, giving your answer to 3 significant figures. 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 block is modelled as a particle, so its weight acts at a single point.

  2. 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.

  3. Resolve perpendicular to the plane to find the normal reaction

    N=mgcos⁡α=12×9.8×2425=112.896 NN=mg\cos\alpha=12\times 9.8\times \frac{24}{25}=112.896\ \text{N}

    On a slope the normal reaction is mgcos⁡αmg\cos\alpha, not mgmg; writing mgmg here is the classic error.

  4. Find the frictional force

    Ffriction=μN=0.3×112.896=33.8688 NF_{\text{friction}}=\mu N=0.3\times 112.896=33.8688\ \text{N}

    Friction opposes the motion, so it always takes energy out of the system.

  5. Find the work done against friction

    Wfriction=μmgdcos⁡α=0.3×12×9.8×16×2425=541.9008 JW_{\text{friction}}=\mu mgd\cos\alpha=0.3\times 12\times 9.8\times 16\times \frac{24}{25}=541.9008\ \text{J}

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

  6. Find the work done by the applied force

    Wdrive=Fd=150×16=2400 JW_{\text{drive}}=Fd=150\times 16=2400\ \text{J}

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

  7. Find the vertical height change

    h=dsin⁡α=16×725=4.48 mh=d\sin\alpha=16\times \frac{7}{25}=4.48\ \text{m}

    It is sin⁡α\sin\alpha, not cos⁡α\cos\alpha, that gives the height gained along a slope.

  8. Find the change in gravitational potential energy

    ΔPE=mgdsin⁡α=12×9.8×16×725=526.848 J\Delta\text{PE}=mgd\sin\alpha=12\times 9.8\times 16\times \frac{7}{25}=526.848\ \text{J}

    The block moves up the slope, so its potential energy increases and the term is positive.

  9. Find the kinetic energy at AA

    12mu2=12×12×02=0 J\tfrac{1}{2}mu^{2}=\tfrac{1}{2}\times 12\times 0^{2}=0\ \text{J}

    This is the kinetic energy the block starts with.

  10. Find the kinetic energy at BB

    12mv2=12×12×v2=6v2 J\tfrac{1}{2}mv^{2}=\tfrac{1}{2}\times 12\times v^{2}=6v^{2}\ \text{J}

    This is the kinetic energy the block finishes with.

  11. Find the change in kinetic energy

    ΔKE=12mv2−12mu2=6v2−0=6v2 J\Delta\text{KE}=\tfrac{1}{2}mv^{2}-\tfrac{1}{2}mu^{2}=6v^{2}-0=6v^{2}\ \text{J}

    A negative value would mean the block has slowed down.

  12. Substitute every term into the work-energy principle

    2400−(541.9008)=6v2+(526.848)2400-\left(541.9008\right)=6v^{2}+\left(526.848\right)

    Each term carries its own sign: work supplied is positive, work done against resistance is subtracted, and a gain in potential energy is added.

  13. Rearrange and solve for the required quantity

    ⇒v=14.9\Rightarrow\quad v=14.9

    This is now the only unknown left in the equation.

  14. Round the answer to 3 significant figures

    v≈14.9 m s−1v\approx14.9\ \text{m s}^{-1}

    The question asks for the answer to 3 significant figures.

  15. Select the option that matches this value

    v=14.9 m s−1v=14.9\ \text{m s}^{-1}

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

Answer
v=14.9 m s−1v=14.9\ \text{m s}^{-1}
Question 3
9 markschallenging
A van of mass 14001400 kg is modelled as a particle. The van moves up a straight road inclined at an angle α\alpha to the horizontal, where sin⁡α=120\sin\alpha=\frac{1}{20}. The resistance to motion is constant and has magnitude 700700 N. The engine of the van works at a constant rate of 27.7227.72 kW. The van is moving at its maximum speed of 2020 m s−1^{-1}. Take g=9.8 m s−2g=9.8\ \text{m s}^{-2}. Select the statement which correctly explains why the driving force is equal to the total resistance to the motion of the van when it is travelling at its maximum speed and which gives the correct magnitude of the driving force.
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 van 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 van 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=1400×9.8×120=686 Nmg\sin\alpha=1400\times 9.8\times \frac{1}{20}=686\ \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

    27.72 kW=27720 W27.72\ \text{kW}=27720\ \text{W}

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

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

    F=Pv=2772020=1386 NF=\frac{P}{v}=\frac{27720}{20}=1386\ \text{N}

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

  8. Check the rate of working against the energy budget

    P−Rv−mgvsin⁡α−mav=27720−14000−13720−0=0P-Rv-mgv\sin\alpha-mav=27720-14000-13720-0=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.

  9. 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.

  10. State the units of the answer

    [F]=N\left[F\right]=\text{N}

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

  11. 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.

  12. 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}.

  13. Recall the formula for gravitational potential energy

    PE=mgh\text{PE}=mgh

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

  14. 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.

  15. Recall the law of friction for a moving body

    Ffriction=μNF_{\text{friction}}=\mu N

    Friction acts along the surface, opposing the direction of motion.

  16. Select the option that matches this value

    F=1386 NF=1386\ \text{N}

    This is the required quantity, with the correct units.

Answer
F=1386 NF=1386\ \text{N}
Question 4
9 markschallenging
A box of mass 88 kg is modelled as a particle. The box moves up a line of greatest slope of a plane inclined at an angle α\alpha to the horizontal, where sin⁡α=35\sin\alpha=\frac{3}{5}, from a point AA to a point BB, where AB=5AB=5 m. The plane is rough and the coefficient of friction between the box and the plane is 0.50.5. A constant force of magnitude 91.291.2 N acts on the box in the direction of motion. The box 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}. Select the statement which correctly explains why the work done against friction is a loss of energy and which gives the correct magnitude of the work done against friction.
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 box is modelled as a particle, so its weight acts at a single point.

  2. List the forces acting on the box

    weight,normal reaction,applied force,resistance\text{weight},\quad \text{normal reaction},\quad \text{applied force},\quad \text{resistance}

    Only forces with a component along the displacement do any work.

  3. 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.

  4. Resolve perpendicular to the plane to find the normal reaction

    N=mgcos⁡α=8×9.8×45=62.72 NN=mg\cos\alpha=8\times 9.8\times \frac{4}{5}=62.72\ \text{N}

    On a slope the normal reaction is mgcos⁡αmg\cos\alpha, not mgmg; writing mgmg here is the classic error.

  5. Find the frictional force

    Ffriction=μN=0.5×62.72=31.36 NF_{\text{friction}}=\mu N=0.5\times 62.72=31.36\ \text{N}

    Friction opposes the motion, so it always takes energy out of the system.

  6. Find the work done against friction

    Wfriction=μmgdcos⁡α=0.5×8×9.8×5×45=156.8 JW_{\text{friction}}=\mu mgd\cos\alpha=0.5\times 8\times 9.8\times 5\times \frac{4}{5}=156.8\ \text{J}

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

  7. Find the work done by the applied force

    Wdrive=Fd=91.2×5=456 JW_{\text{drive}}=Fd=91.2\times 5=456\ \text{J}

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

  8. Find the vertical height change

    h=dsin⁡α=5×35=3 mh=d\sin\alpha=5\times \frac{3}{5}=3\ \text{m}

    It is sin⁡α\sin\alpha, not cos⁡α\cos\alpha, that gives the height gained along a slope.

  9. Find the change in gravitational potential energy

    ΔPE=mgdsin⁡α=8×9.8×5×35=235.2 J\Delta\text{PE}=mgd\sin\alpha=8\times 9.8\times 5\times \frac{3}{5}=235.2\ \text{J}

    The box moves up the slope, so its potential energy increases and the term is positive.

  10. Find the kinetic energy at AA

    12mu2=12×8×02=0 J\tfrac{1}{2}mu^{2}=\tfrac{1}{2}\times 8\times 0^{2}=0\ \text{J}

    This is the kinetic energy the box starts with.

  11. Find the kinetic energy at BB

    12mv2=12×8×42=64 J\tfrac{1}{2}mv^{2}=\tfrac{1}{2}\times 8\times 4^{2}=64\ \text{J}

    This is the kinetic energy the box finishes with.

  12. Find the change in kinetic energy

    ΔKE=12mv2−12mu2=64−0=64 J\Delta\text{KE}=\tfrac{1}{2}mv^{2}-\tfrac{1}{2}mu^{2}=64-0=64\ \text{J}

    A negative value would mean the box has slowed down.

  13. Substitute every term into the work-energy principle

    456−(156.8)=64+(235.2)456-\left(156.8\right)=64+\left(235.2\right)

    Each term carries its own sign: work supplied is positive, work done against resistance is subtracted, and a gain in potential energy is added.

  14. Rearrange and solve for the required quantity

    ⇒Wfriction=156.8\Rightarrow\quad W_{\text{friction}}=156.8

    This is now the only unknown left in the equation.

  15. Check that the energy budget balances

    Wdrive−Wresistance−ΔKE−ΔPE=456−156.8−64−235.2=0W_{\text{drive}}-W_{\text{resistance}}-\Delta\text{KE}-\Delta\text{PE}=456-156.8-64-235.2=0

    Every joule supplied is accounted for, so the answer is consistent.

  16. Cross-check with Newton's second law

    Fresultant=12.8 N  ⟹  a=12.88=1.6 m s−2F_{\text{resultant}}=12.8\ \text{N}\;\Longrightarrow\;a=\frac{12.8}{8}=1.6\ \text{m s}^{-2}

    The resultant force is the applied force minus the resistance and minus the component of the weight along the motion.

  17. Confirm the speeds with v2=u2+2adv^{2}=u^{2}+2ad

    v2=02+2×1.6×5=16v^{2}=0^{2}+2\times 1.6\times 5=16

    This independent route reproduces the same motion, confirming the energy calculation.

  18. Select the option that matches this value

    Wfriction=156.8 JW_{\text{friction}}=156.8\ \text{J}

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

Answer
Wfriction=156.8 JW_{\text{friction}}=156.8\ \text{J}
Question 5
9 markschallenging
A car of mass 10001000 kg is modelled as a particle. The car 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. The engine of the car works at a constant rate of 1818 kW. The car is moving at its maximum speed of 2020 m s−1^{-1}. Take g=9.8 m s−2g=9.8\ \text{m s}^{-2}. Find the magnitude of the resistance to motion of the car.
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 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. Use the condition for maximum speed

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

    At maximum speed the car 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=1000×9.8×114=700 Nmg\sin\alpha=1000\times 9.8\times \frac{1}{14}=700\ \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

    18 kW=18000 W18\ \text{kW}=18000\ \text{W}

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

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

    F=Pv=1800020=900 NF=\frac{P}{v}=\frac{18000}{20}=900\ \text{N}

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

  8. Solve the equation of motion for the resistance

    R=F−mgsin⁡α−ma=200 NR=F-mg\sin\alpha-ma=200\ \text{N}

    The driving force balances the resistance, the component of the weight along the road and the force needed to accelerate the car.

  9. Check the rate of working against the energy budget

    P−Rv−mgvsin⁡α−mav=18000−4000−14000−0=0P-Rv-mgv\sin\alpha-mav=18000-4000-14000-0=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

    [R]=N\left[R\right]=\text{N}

    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. Recall the law of friction for a moving body

    Ffriction=μNF_{\text{friction}}=\mu N

    Friction acts along the surface, opposing the direction of motion.

  17. State the final answer

    R=200 NR=200\ \text{N}

    This is the required quantity, with the correct units.

Answer
R=200 NR=200\ \text{N}

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