FM1-2 Work, energy and power — revision question pack

1 specification points · notes, questions, answers and worked methods

Checked against Edexcel 9FM0 section FM1-2. Review basis: the qualification registry sourced from the Pearson Edexcel Level 3 Advanced GCE in Further Mathematics (9FM0) specification; registry verification recorded 17 July 2026.

How this checking works

FM1-2.1 · Kinetic and potential energy, work and power. The work-energy principle. The principle of conservation of mechanical energy.

Explanation

Kinetic energy is $\tfrac12mv^2$ , while gravitational potential energy changes by $mg\Delta h$ . Work is force times displacement for a constant parallel force and $\int F\,dx$ when the force varies with position.
The work-energy principle states that work done by the resultant force equals the change in kinetic energy.
Mechanical energy is conserved only when no non-conservative force does work; driving forces and resistance must otherwise appear through their work.
Power is $P=dW/dt$ and equals $Fv$ only for the component of force parallel to velocity.
Examiners expect a consistent energy balance, correct signs for work and resolved forces on inclined planes.

A body moving up an incline with driving force, resistance and weight shown.

Worked example

A $900\,\text{kg}$ car travels up a slope of angle $4^\circ$ at constant speed $18\,\text{m s}^{-1}$ . Resistance is $350\,\text{N}$ . Taking $g=9.8\,\text{m s}^{-2}$ , find the engine power.

1.Constant speed gives zero resultant force along the slope.
2.The driving force is $F=900(9.8)\sin4^\circ+350=965.2\,\text{N}$ to the stated accuracy.
3. $P=Fv=965.2(18)=1.737\times10^4\,\text{W}$ .

Answer: The engine power is $17.4\,\text{kW}$ to $3$ significant figures.

Common mistakes

Don't use $mg\cos\alpha$ instead of $mg\sin\alpha$ for the component of weight down the slope.
Don't conserve mechanical energy despite a stated resistance force.
Don't use the full force magnitude in $P=Fv$ when force and velocity are not parallel.

Exam tip

State whether the equation is work-energy, conservation of energy or power before substituting values.

Tier 1 · Easy

1.
A $1200\,\text{kg}$ car increases its speed from $10\,\text{m s}^{-1}$ to $14\,\text{m s}^{-1}$ . Calculate the net work done on the car.
(2)
(Total for Question 1 is 2 marks)
2.
A $2\,\text{kg}$ particle is released from rest on a smooth slope. Find its speed after it has fallen through a vertical distance of $5\,\text{m}$ . Take $g=9.8\,\text{m s}^{-2}$ .
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
A $1200\,\text{kg}$ car climbs a road inclined at $5^\circ$ to the horizontal at a constant speed of $20\,\text{m s}^{-1}$ . The resistance is $400\,\text{N}$ . Taking $g=9.8\,\text{m s}^{-2}$ , find the engine power.
(4)
(Total for Question 1 is 4 marks)
2.
The angle $\alpha$ of a slope satisfies $\sin\alpha=1/20$ . A $500\,\text{kg}$ trolley is driven $80\,\text{m}$ uphill at constant speed against a resistance modelled as a constant force of $150\,\text{N}$ . Use $g=9.8\,\text{m s}^{-2}$ . Determine the work done by the engine and state one limitation of modelling the resistance as constant.
(5)
(Total for Question 2 is 5 marks)
3.
A particle of mass $4\,\text{kg}$ starts from rest and moves up a plane inclined at an angle $\alpha$ , where $\sin\alpha=1/4$ . A constant force of $50\,\text{N}$ acts on the particle up the line of greatest slope. A constant non-gravitational resistance of magnitude $8\,\text{N}$ opposes the motion. The particle moves $5\,\text{m}$ up the plane. Taking $g=9.8\,\text{m s}^{-2}$ , find the work done by the applied force, the gain in gravitational potential energy, the work done against the resistance and the particle's final speed.
(5)
(Total for Question 3 is 5 marks)

Tier 3 · Hard

1.
A $2\,\text{kg}$ particle moves along a horizontal line. A constant driving force of $20\,\text{N}$ acts forwards while the resistance after displacement $x\,\text{m}$ is $(2+0.5x)\,\text{N}$ . Its speed at $x=0$ is $3\,\text{m s}^{-1}$ . Find its speed when $x=8$ .
(5)
(Total for Question 1 is 5 marks)
2.
A $2\,\text{kg}$ particle is projected at $\sqrt{14}\,\text{m s}^{-1}$ up a plane inclined at $\alpha$ to the horizontal, where $\sin\alpha=1/21$ . Non-gravitational forces are modelled as a resistance of magnitude $(1+3x^2)\,\text{N}$ , where $x\,\text{m}$ is the particle's displacement up the plane. Use $g=9.8\,\text{m s}^{-2}$ . Determine the distance $s\,\text{m}$ travelled before the particle first comes to rest, showing that it is the only possible positive value of $s$ .
(6)
(Total for Question 2 is 6 marks)
3.
An $800\,\text{kg}$ vehicle moves along a horizontal road at speed $v\,\text{m s}^{-1}$ , where $v>0$ . The resistance is modelled as a force of magnitude $20v\,\text{N}$ . The engine of the vehicle is working at a constant rate of $48\,\text{kW}$ . Determine the complete range of speeds for which the vehicle has an instantaneous acceleration of at least $0.5\,\text{m s}^{-2}$ .
(6)
(Total for Question 3 is 6 marks)
4.
A $2\,\text{kg}$ particle moves along a horizontal line. At displacement $x\,\text{m}$ from its initial position, a driving force of magnitude $(18-\tfrac12x^2)\,\text{N}$ acts in the direction of motion and a constant resistance of magnitude $3\,\text{N}$ opposes the motion. The particle starts at $x=0$ with speed $5\,\text{m s}^{-1}$ . Find its speed at $x=4$ . Find also the displacement at which its speed is greatest, the greatest speed and the instantaneous power supplied by the driving force there.
(7)
(Total for Question 4 is 7 marks)
5.
A box of mass $5\,\text{kg}$ is pulled from speed $2\,\text{m s}^{-1}$ along a rough horizontal floor by a constant tension of $26\,\text{N}$ in a rope inclined at an angle $\theta$ above the horizontal, where $\sin\theta=5/13$ . The coefficient of friction between the box and the floor is $1/3$ . The box moves $4\,\text{m}$ without losing contact with the floor. Taking $g=9.8\,\text{m s}^{-2}$ , find the work done by the tension, the work done against friction and the final speed of the box. Find also the instantaneous power supplied by the tension at the final speed.
(7)
(Total for Question 5 is 7 marks)

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

FM1-2.1 · Kinetic and potential energy, work and power. The work-energy principle. The principle of conservation of mechanical energy.

Tier 1 · Easy

Mark scheme for FM1-2.1 Tier 1 · Easy
Question	Scheme	Marks
1	$57\,600\,\text{J}$	2
(2 marks)		2
Notes
Net work equals the gain in kinetic energy: $W=\tfrac12(1200)(14^2-10^2)=600(96)=57\,600\,\text{J}$ .
2	Mechanical energy is conserved because the slope is smooth The loss of gravitational potential energy is $2(9.8)(5)=98\,\text{J}$ $98=\tfrac12(2)v^2$ , so $v=7\sqrt2\,\text{m s}^{-1}$	3
(3 marks)		3
Notes
Mechanical energy is conserved because the slope is smooth. The loss of gravitational potential energy is $2(9.8)(5)=98\,\text{J}$ , so $98=\tfrac12(2)v^2$ . Hence $v=\sqrt{98}=7\sqrt2\,\text{m s}^{-1}$ .

Tier 2 · Standard

Mark scheme for FM1-2.1 Tier 2 · Standard
Question	Scheme	Marks
1	$2.85\times10^4\,\text{W}$ , or $28.5\,\text{kW}$ (3 s.f.)	4
(4 marks)		4
Notes
Constant speed gives zero resultant force along the slope. The driving force is $F=1200(9.8)\sin5^\circ+400=1424.9\,\text{N}$ . Therefore $P=Fv=1424.9(20)=2.85\times10^4\,\text{W}$ .
2	Constant speed means that the kinetic-energy change is zero The gain in gravitational potential energy is $19\,600\,\text{J}$ The work done against resistance is $12\,000\,\text{J}$ The work done by the engine is $31\,600\,\text{J}$ For example, the model ignores changes in resistance caused by speed, the surface or the surroundings	5
(5 marks)		5
Notes
Constant speed means that the kinetic-energy change is zero. The gain in gravitational potential energy is $500(9.8)(80)(1/20)=19\,600\,\text{J}$ , and the work done against resistance is $150(80)=12\,000\,\text{J}$ . Therefore the engine does $19\,600+12\,000=31\,600\,\text{J}$ of work. A constant-resistance model neglects factors such as changes in speed, surface conditions or wind.
3	The work done by the applied force is $50(5)=250\,\text{J}$ The gain in gravitational potential energy is $4(9.8)(5)(1/4)=49\,\text{J}$ The work done against the resistance is $8(5)=40\,\text{J}$ Work-energy gives $\tfrac12(4)v^2=250-49-40=161$ Hence $v^2=80.5$ and the final speed is $8.97\,\text{m s}^{-1}$ to $3$ significant figures	5
(5 marks)		5
Notes
The applied force does $50(5)=250\,\text{J}$ of work. The gain in gravitational potential energy is $4(9.8)(5)(1/4)=49\,\text{J}$ , and $8(5)=40\,\text{J}$ is dissipated against the resistance. Since the particle starts from rest, work-energy gives $\tfrac12(4)v^2=250-49-40=161$ . Thus $v^2=80.5$ and $v=\sqrt{80.5}=8.972179\ldots\,\text{m s}^{-1}$ , so the final speed is $8.97\,\text{m s}^{-1}$ to $3$ significant figures.

Tier 3 · Hard

Mark scheme for FM1-2.1 Tier 3 · Hard
Question	Scheme	Marks
1	$\sqrt{137}\,\text{m s}^{-1}\approx11.7\,\text{m s}^{-1}$	5
(5 marks)		5
Notes
The driving work is $20(8)=160\,\text{J}$ . The work against resistance is $\int_0^8(2+0.5x)\,dx=[2x+0.25x^2]_0^8=32\,\text{J}$ . Hence the net work is $128\,\text{J}$ . By work-energy, $\tfrac12(2)v^2-\tfrac12(2)(3^2)=128$ , so $v^2=137$ and $v=\sqrt{137}\,\text{m s}^{-1}$ .
2	The downslope component of weight is $14/15\,\text{N}$ The total opposing force is $(29/15+3x^2)\,\text{N}$ The initial kinetic energy is $14\,\text{J}$ The stopping equation is $s^3+29s/15=14$ $s=2.14\,\text{m}$ to $3$ significant figures This is the only positive solution because $3s^2+29/15>0$	6
(6 marks)		6
Notes
The downslope component of weight is $2(9.8)(1/21)=14/15\,\text{N}$ , so the total opposing force is $29/15+3x^2$ . The initial kinetic energy is $\tfrac12(2)(\sqrt{14})^2=14\,\text{J}$ . If the stopping distance is $s$ , work-energy gives $\int_0^s(29/15+3x^2)\,dx=14$ , hence $s^3+29s/15=14$ . Solving numerically gives $s=2.143970\ldots$ , so $s=2.14\,\text{m}$ to $3$ significant figures. Since the derivative of $s^3+29s/15$ is $3s^2+29/15>0$ , there is only one real, and hence only one positive, solution.
3	The driving force is $48\,000/v\,\text{N}$ An acceleration of at least $0.5\,\text{m s}^{-2}$ requires $48\,000/v-20v\ge400$ Since $v>0$ , this is equivalent to $20v^2+400v-48\,000\le0$ Dividing by $20$ gives $v^2+20v-2400\le0$ The inequality factors as $(v-40)(v+60)\le0$ Combining $-60\le v\le40$ with $v>0$ gives $0<v\le40\,\text{m s}^{-1}$	6
(6 marks)		6
Notes
The driving force at speed $v$ is $P/v=48\,000/v$ , so the resultant force is $48\,000/v-20v$ . Requiring this to be at least $800(0.5)=400\,\text{N}$ and multiplying by the positive speed gives $v^2+20v-2400\le0$ . This is $(v-40)(v+60)\le0$ , so the physical restriction $v>0$ leaves $0<v\le40\,\text{m s}^{-1}$ .
4	The net work from $0$ to $x$ is $\int_0^x(15-\tfrac12s^2)\,ds=15x-x^3/6$ Work-energy gives $v^2=25+15x-x^3/6$ At $x=4$ , $v^2=223/3$ , so $v=\sqrt{223/3}\,\text{m s}^{-1}$ The speed is greatest when the resultant force is zero: $15-x^2/2=0$ Hence the displacement is $x=\sqrt{30}\,\text{m}$ The greatest speed is $\sqrt{25+10\sqrt{30}}\,\text{m s}^{-1}$ The driving force is then $3\,\text{N}$ , so its power is $3\sqrt{25+10\sqrt{30}}\,\text{W}$	7
(7 marks)		7
Notes
The resultant force is $15-x^2/2$ , so its work from $0$ to $x$ is $15x-x^3/6$ . Since the mass is $2\,\text{kg}$ , the kinetic energy is $v^2$ , giving $v^2=25+15x-x^3/6$ . Substitution at $x=4$ gives $223/3$ . Speed increases while the resultant force is positive and decreases after it becomes negative, so the maximum is at $x=\sqrt{30}$ . Substitution gives the greatest speed. At that point the driving force equals the $3\,\text{N}$ resistance, and $P=Fv$ gives the stated power.
5	Vertical equilibrium gives $R=5(9.8)-26(5/13)=39\,\text{N}$ The friction force has magnitude $(1/3)(39)=13\,\text{N}$ The work done by the tension is $26(4)(12/13)=96\,\text{J}$ The work done against friction is $13(4)=52\,\text{J}$ Work-energy gives $\tfrac12(5)v^2-\tfrac12(5)(2^2)=96-52$ Hence $v^2=108/5$ , so the final speed is $6\sqrt{15}/5\,\text{m s}^{-1}$ The instantaneous power is $26v\cos\theta=24v=144\sqrt{15}/5\,\text{W}$	7
(7 marks)		7
Notes
The upward component of the tension reduces the normal reaction to $R=49-10=39\,\text{N}$ , so the friction force is $13\,\text{N}$ . Since $\cos\theta=12/13$ , the tension does $26(4)(12/13)=96\,\text{J}$ of work, while $52\,\text{J}$ is dissipated against friction. Work-energy therefore gives $(5/2)v^2-10=44$ , so $v^2=108/5$ and $v=6\sqrt{15}/5\,\text{m s}^{-1}$ . At that speed the component of tension parallel to the velocity is $26(12/13)=24\,\text{N}$ , giving power $24v=144\sqrt{15}/5\,\text{W}$ .