6.02a Work done: concept and definition

1 A bus of mass 8 tonnes is driven up a hill on a straight road. On one part of the hill, the power of the driving force on the bus is constant at 20 kW for one minute.

1. Calculate how much work is done by the driving force in this time. During this minute the speed of the bus increases from \(8 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) to \(12 \mathrm {~ms} ^ { - 1 }\) and, in addition to the work done against gravity, 125000 J of work is done against the resistance to motion of the bus parallel to the slope.
2. Calculate the change in the kinetic energy of the bus.
3. Calculate the vertical displacement of the bus. On another stretch of the road, a driving force of power 26 kW is required to propel the bus up a slope of angle \(\theta\) to the horizontal at a constant speed of \(6.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\), against a resistance to motion of 225 N parallel to the slope.
4. Calculate the angle \(\theta\). The bus later travels up the same slope of angle \(\theta\) to the horizontal at the same constant speed of \(6.5 \mathrm {~ms} ^ { - 1 }\) but now against a resistance to motion of 155 N parallel to the slope.
5. Calculate the power of the driving force on the bus.

2 This question is about 'kart gravity racing' in which, after an initial push, unpowered home-made karts race down a sloping track. The moving karts have only the following resistive forces and these both act in the direction opposite to the motion.

• A force \(R\), called rolling friction, with magnitude \(0.01 M g \cos \theta \mathrm {~N}\) where \(M \mathrm {~kg}\) is the mass of the kart and driver and \(\theta\) is the angle of the track with the horizontal
• A force \(F\) of varying magnitude, due to air resistance

A kart with its driver has a mass of 80 kg .
One stretch of track slopes uniformly downwards at \(4 ^ { \circ }\) to the horizontal. The kart travels 12 m down this stretch of track. The total work done by the kart against both rolling friction and air resistance is 455 J .

1. Calculate the work done against air resistance.
2. During this motion, the kart's speed increases from \(2 \mathrm {~ms} ^ { - 1 }\) to \(v \mathrm {~ms} ^ { - 1 }\). Use an energy method to calculate \(v\). To reach the starting line, the kart (with the driver seated) is pushed up a slope against rolling friction and air resistance. At one point the slope is at \(5 ^ { \circ }\) to the horizontal, the air resistance is 15 N , the acceleration of the kart is \(1.5 \mathrm {~m} \mathrm {~s} ^ { - 2 }\) up the slope and the power of the pushing force is 405 W .
3. Calculate the speed of the kart at this point.

2 A car of mass 850 kg is travelling along a road that is straight but not level.
On one section of the road the car travels at constant speed and gains a vertical height of 60 m in 20 seconds. Non-gravitational resistances to its motion (e.g. air resistance) are negligible.

1. Show that the average power produced by the car is about 25 kW . On a horizontal section of the road, the car develops a constant power of exactly 25 kW and there is a constant resistance of 800 N to its motion.
2. Calculate the maximum possible steady speed of the car.
3. Find the driving force and acceleration of the car when its speed is \(10 \mathrm {~ms} ^ { - 1 }\). When travelling along the horizontal section of road, the car accelerates from \(15 \mathrm {~ms} ^ { - 1 }\) to \(20 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) in 6.90 seconds with the same constant power and constant resistance.
4. By considering work and energy, find how far the car travels while it is accelerating. When the car is travelling at \(20 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) up a constant slope inclined at \(\arcsin ( 0.05 )\) to the horizontal, the driving force is removed. Subsequently, the resistance to the motion of the car remains constant at 800 N .
5. What is the speed of the car when it has travelled a further 105 m up the slope?

3

1. A car of mass 900 kg is travelling at a steady speed of \(16 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) up a hill inclined at arcsin 0.1 to the horizontal. The power required to do this is 20 kW . Calculate the resistance to the motion of the car.
2. A small box of mass 11 kg is placed on a uniform rough slope inclined at arc \(\cos \frac { 12 } { 13 }\) to the horizontal. The coefficient of friction between the box and the slope is \(\mu\).
   1. Show that if the box stays at rest then \(\mu \geqslant \frac { 5 } { 12 }\). For the remainder of this question, the box moves on a part of the slope where \(\mu = 0.2\).
      The box is projected up the slope from a point P with an initial speed of \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\). It travels a distance of 1.5 m along the slope before coming instantaneously to rest. During this motion, the work done against air resistance is 6 joules per metre.
   2. Calculate the value of \(v\). As the box slides back down the slope, it passes through its point of projection P and later reaches its initial speed at a point Q . During this motion, once again the work done against air resistance is 6 joules per metre.
   3. Calculate the distance PQ.

4 Jack and Jill are raising a pail of water vertically using a light inextensible rope. The pail and water have total mass 20 kg . In parts (i) and (ii), all non-gravitational resistances to motion may be neglected.

1. How much work is done to raise the pail from rest so that it is travelling upwards at \(0.5 \mathrm {~ms} ^ { - 1 }\) when at a distance of 4 m above its starting position?
2. What power is required to raise the pail at a steady speed of \(0.5 \mathrm {~ms} ^ { - 1 }\) ? Jack falls over and hurts himself. He then slides down a hill.
   His mass is 35 kg and his speed increases from \(1 \mathrm {~ms} ^ { - 1 }\) to \(3 \mathrm {~ms} ^ { - 1 }\) while descending through a vertical height of 3 m .
3. How much work is done against friction? In Jack's further motion, he slides down a slope at an angle \(\alpha\) to the horizontal where \(\sin \alpha = 0.1\). The frictional force on him is now constant at 150 N . For this part of the motion, Jack's initial speed is \(3 \mathrm {~ms} ^ { - 1 }\).
4. How much further does he slide before coming to rest?

2

1. A small block of mass 25 kg is on a long, horizontal table. Each side of the block is connected to a small sphere by means of a light inextensible string passing over a smooth pulley. Fig. 2 shows this situation. Sphere A has mass 5 kg and sphere B has mass 20 kg . Each of the spheres hangs freely. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{81efb50d-c89d-4ce1-94d7-592c946f6176-3_487_1123_466_552} \captionsetup{labelformat=empty} \caption{Fig. 2}
   \end{figure} Initially the block moves on a smooth part of the table. With the block at a point O , the system is released from rest with both strings taut.
   1. (A) Is mechanical energy conserved in the subsequent motion? Give a brief reason for your answer.
      (B) Why is no work done by the block against the reaction of the table on it? The block reaches a speed of \(1.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at point P .
   2. Use an energy method to calculate the distance OP. The block continues moving beyond P , at which point the table becomes rough. After travelling two metres beyond P , the block passes through point Q . The block does 180 J of work against resistances to its motion from P to Q .
   3. Use an energy method to calculate the speed of the block at Q .
2. A tree trunk of mass 450 kg is being pulled up a slope inclined at \(20 ^ { \circ }\) to the horizontal. Calculate the power required to pull the trunk at a steady speed of \(2.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) against a frictional force of 2000 N .

4 A box of mass 16 kg is on a uniformly rough horizontal floor with an applied force of fixed direction but varying magnitude \(P\) N acting as shown in Fig. 4. You may assume that the box does not tip for any value of \(P\). The coefficient of friction between the box and the floor is \(\mu\). \begin{figure}[h] \end{figure} Initially the box is at rest and on the point of slipping with \(P = 58\).

1. Show that \(\mu\) is about 0.25 . In the rest of this question take \(\mu\) to be exactly 0.25 .
   The applied force on the box is suddenly increased so that \(P = 70\) and the box moves against friction with the floor and another horizontal retarding force, \(S\). The box reaches a speed of \(1.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) from rest after 5 seconds; during this time the box slides 3 m .
2. Calculate the work done by the applied force of 70 N and also the average power developed by this force over the 5 seconds.
3. By considering the values of time, distance and speed, show that an object starting from rest that travels 3 m while reaching a speed of \(1.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) in 5 seconds cannot be moving with constant acceleration. The reason that the acceleration is not constant is that the retarding force \(S\) is not constant.
4. Calculate the total work done by the retarding force \(S\).

2. A pump raises water from a well 12 metres below the ground and ejects the water through a pipe of diameter 10 cm at a speed of \(6 \mathrm {~ms} ^ { - 1 }\). Given that the mass of \(1 \mathrm {~m} ^ { 3 }\) of water is 1000 kg ,

1. find, in terms of \(\pi\), the mass of water discharged by the pipe every second,
2. find in kJ , correct to 3 significant figures, the total mechanical energy gained by the water per second.

7 A particle \(P\) of mass 0.2 kg is released from rest at a point \(O\) and falls vertically. Air resistance of magnitude \(\frac { v ^ { 2 } } { 2000 } \mathrm {~N}\) acts upwards on \(P\), where \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) is the velocity of \(P\) when it has fallen a distance of \(x \mathrm {~m}\).

1. Show that \(\left( \frac { 400 v } { 3920 - v ^ { 2 } } \right) \frac { \mathrm { d } v } { \mathrm {~d} x } = 1\).
2. Find \(v ^ { 2 }\) in terms of \(x\) and hence show that \(v ^ { 2 } < 3920\) for all values of \(x\).
3. Find the work done against the air resistance while \(P\) is falling, from \(O\), to the point where its downward acceleration is \(5.8 \mathrm {~m} \mathrm {~s} ^ { - 2 }\). }{www.ocr.org.uk}) after the live examination series.
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5. A rocket is launched vertically upwards under gravity from rest at time \(t = 0\). The rocket propels itself upward by ejecting burnt fuel vertically downwards at a constant speed \(u\) relative to the rocket. The initial mass of the rocket, including fuel, is \(M\). At time \(t\), before all the fuel has been used up, the mass of the rocket, including fuel, is \(M ( 1 - k t )\) and the speed of the rocket is \(v\).

1. Show that \(\frac { \mathrm { d } v } { \mathrm {~d} t } = \frac { k u } { 1 - k t } - g\).
2. Hence find the speed of the rocket when \(t = \frac { 1 } { 3 k }\).

1. Two constant forces \(\mathbf { F } _ { 1 }\) and \(\mathbf { F } _ { 2 }\) are the only forces acting on a particle. \(\mathbf { F } _ { 1 }\) has magnitude 9 N and acts in the direction of \(2 \mathbf { i } + \mathbf { j } + 2 \mathbf { k } . \mathbf { F } _ { 2 }\) has magnitude 18 N and acts in the direction of \(\mathbf { i } + 8 \mathbf { j } - 4 \mathbf { k }\).

Find the total work done by the two forces in moving the particle from the point with position vector \(( \mathbf { i } + \mathbf { j } + \mathbf { k } ) \mathrm { m }\) to the point with position vector \(( 3 \mathbf { i } + 2 \mathbf { j } - \mathbf { k } ) \mathrm { m }\).
(Total 6 marks)

6. A rocket-driven car moves along a straight horizontal road. The car has total initial mass \(M\). It propels itself forwards by ejecting mass backwards at a constant rate \(\lambda\) per unit time at a constant speed \(U\) relative to the car. The car starts from rest at time \(t = 0\). At time \(t\) the speed of the car is \(v\). The total resistance to motion is modelled as having magnitude \(k v\), where \(k\) is a constant. Given that \(t < \frac { M } { \lambda }\), show that

1. \(\frac { \mathrm { d } v } { \mathrm {~d} t } = \frac { \lambda U - k v } { M - \lambda t }\),
2. \(v = \frac { \lambda U } { k } \left\{ 1 - \left( 1 - \frac { \lambda t } { M } \right) ^ { \frac { k } { \lambda } } \right\}\).
   (6)
   (Total 13 marks)

1. A bead of mass 0.5 kg is threaded on a smooth straight wire. The only forces acting on the bead are a constant force ( \(4 \mathbf { i } + 7 \mathbf { j } + 2 \mathbf { k }\) ) N and the normal reaction of the wire. The bead starts from rest at the point \(A\) with position vector \(( \mathbf { i } + 2 \mathbf { j } + 3 \mathbf { k } ) \mathrm { m }\) and moves to the point \(B\) with position vector \(( 4 \mathbf { i } + 3 \mathbf { j } - 2 \mathbf { k } ) \mathrm { m }\).

Find the speed of the bead when it reaches \(B\).
(4)

7. A motor boat of mass \(M\) is moving in a straight line, with its engine switched off, across a stretch of still water. The boat is moving with speed \(U\) when, at time \(t = 0\), it develops a leak. The water comes in at a constant rate so that at time \(t\), the mass of water in the boat is \(\lambda t\). At time \(t\) the speed of the boat is \(v\) and it experiences a total resistance to motion of magnitude \(2 \lambda v\).

1. Show that \(( M + \lambda t ) \frac { \mathrm { d } v } { \mathrm {~d} t } + 3 \lambda v = 0\).
   (6)
2. Show that the time taken for the speed of the boat to reduce to \(\frac { 1 } { 2 } U\) is \(\frac { M } { \lambda } \left( 2 ^ { \frac { 1 } { 3 } } - 1 \right)\).
   (6) The boat sinks when the mass of water inside the boat is \(M\).
3. Show that the boat does not sink before the speed of the boat is \(\frac { 1 } { 2 } U\).

1. \hspace{0pt} [In this question \(\mathbf { i }\) and \(\mathbf { j }\) are horizontal unit vectors.]

A small bead of mass 0.5 kg is threaded on a smooth horizontal wire. The bead is initially at rest at the point with position vector \(( \mathbf { i } - 6 \mathbf { j } ) \mathrm { m }\). A constant horizontal force \(\mathbf { P } \mathrm { N }\) then acts on the bead causing it to move along the wire. The bead passes through the point with position vector ( \(7 \mathbf { i } - 14 \mathbf { j }\) ) m with speed \(2 \sqrt { 7 } \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Given that \(\mathbf { P }\) is parallel to ( \(6 \mathbf { i } + \mathbf { j }\) ), find \(\mathbf { P }\).
(6)

1 A child is pulling a toy block in a straight line along a horizontal floor.
The block is moving with a constant speed of \(2 \mathrm {~ms} ^ { - 1 }\) by means of a constant force of magnitude 20 N acting at an angle of \(\theta ^ { \circ }\) above the horizontal. The work done by the force in 10 s is 350 J . Calculate the value of \(\theta\).

5 In the diagram below, points \(\mathrm { A } , \mathrm { B }\) and C lie in the same vertical plane. The slope AB is inclined at an angle of \(30 ^ { \circ }\) to the horizontal and \(\mathrm { AB } = 5 \mathrm {~m}\). The point B is a vertical distance of 6.5 m above horizontal ground. The point C lies on the horizontal ground. \includegraphics[max width=\textwidth, alt={}, center]{a96a0ebe-8f4f-4d79-9d11-9d348ef72314-6_601_1285_395_244} Starting at A , a particle P , of mass \(m \mathrm {~kg}\), moves along the slope towards B , under the action of a constant force \(\mathbf { F }\). The force \(\mathbf { F }\) has a magnitude of 50 N and acts at an angle of \(\theta ^ { \circ }\) to AB in the same vertical plane as A and B . When P reaches \(\mathrm { B } , \mathbf { F }\) is removed, and P moves under gravity landing at C . It is given that

• the speed of P at A is \(3 \mathrm {~m} \mathrm {~s} ^ { - 1 }\),
• the speed of P at B is \(6 \mathrm {~ms} ^ { - 1 }\),
• the speed of P at C is \(12 \mathrm {~m} \mathrm {~s} ^ { - 1 }\),
• 58 J of work is done against non-gravitational resistances as P moves from A to B ,
• 42 J of work is done against non-gravitational resistances as P moves from B to C .
  1. By considering the motion from B to C, show that \(m = 4.33\) correct to 3 significant figures.
  2. By considering the motion from A to B , determine the value of \(\theta\).
  3. Calculate the power of \(\mathbf { F }\) at the instant that P reaches B .

1 A clock is driven by a 5 kg sphere falling once through a vertical distance of 120 cm over 2 days. Calculate, in watts, the average power developed by the falling sphere.

7 A box B of mass \(m \mathrm {~kg}\) is raised vertically by an engine working at a constant rate of \(k m g \mathrm {~W}\). Initially B is at rest. The speed of B when it has been raised a distance \(x \mathrm {~m}\) is denoted by \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\).

1. Show that \(v ^ { 2 } \frac { d v } { d x } = ( k - v ) g\).
2. Verify that \(\mathrm { gx } = \mathrm { k } ^ { 2 } \ln \left( \frac { \mathrm { k } } { \mathrm { k } - \mathrm { v } } \right) - \mathrm { kv } - \frac { 1 } { 2 } \mathrm { v } ^ { 2 }\).
3. By using the work-energy principle, show that the time taken for B to reach a speed \(V \mathrm {~m} \mathrm {~s} ^ { - 1 }\) from rest is given by \(\frac { \mathrm { k } } { \mathrm { g } } \ln \left( \frac { \mathrm { k } } { \mathrm { k } - \mathrm { V } } \right) - \frac { \mathrm { V } } { \mathrm { g } }\).

2. \begin{figure}[h] \end{figure} Figure 1 shows a ramp inclined at an angle \(\theta\) to the horizontal, where \(\sin \theta = \frac { 2 } { 7 }\) A parcel of mass 4 kg is projected, with speed \(5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\), from a point \(A\) on the ramp.
The parcel moves up a line of greatest slope of the ramp and first comes to instantaneous rest at the point \(B\), where \(A B = 2.5 \mathrm {~m}\).
The parcel is modelled as a particle.
The total resistance to the motion of the parcel from non-gravitational forces is modelled as a constant force of magnitude \(R\) newtons.

1. Use the work-energy principle to show that \(R = 8.8\) After coming to instantaneous rest at \(B\), the parcel slides back down the ramp. The total resistance to the motion of the particle is modelled as a constant force of magnitude 8.8N.
2. Find the speed of the parcel at the instant it returns to \(A\).
3. Suggest two improvements that could be made to the model.

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1. \begin{figure}[h] \end{figure} A small book of mass \(m\) is held on a rough straight desk lid which is inclined at an angle \(\alpha\) to the horizontal, where \(\tan \alpha = \frac { 3 } { 4 }\). The book is released from rest at a distance of 0.5 m from the edge of the desk lid, as shown in Figure 1. The book slides down the desk lid and then hits the floor that is 0.8 m below the edge of the desk lid. The coefficient of friction between the book and the desk lid is 0.4 The book is modelled as a particle which, after leaving the desk lid, is assumed to move freely under gravity.

1. Find, in terms of \(m\) and \(g\), the magnitude of the normal reaction on the book as it slides down the desk lid.
2. Use the work-energy principle to find the speed of the book as it hits the floor.