6.02b Calculate work: constant force, resolved component

149 questions

Sort by: Default | Easiest first | Hardest first
WJEC Further Unit 3 2024 June Q3
5 marks Moderate -0.8
3. Three forces \(( 4 \mathbf { i } - 7 \mathbf { j } + 9 \mathbf { k } ) \mathrm { N } , ( 5 \mathbf { i } + 3 \mathbf { j } - 8 \mathbf { k } ) \mathrm { N }\) and \(( - 2 \mathbf { i } + 6 \mathbf { j } - 11 \mathbf { k } ) \mathrm { N }\) act on a particle.
  1. Find the resultant \(\mathbf { R }\) of the three forces.
  2. The points \(A\) and \(B\) have position vectors \(( 3 \mathbf { i } + 4 \mathbf { j } - 12 \mathbf { k } ) \mathrm { m }\) and \(( a \mathbf { i } + 7 \mathbf { j } - 10 \mathbf { k } ) \mathrm { m }\) respectively, where \(a\) is a constant. The work done by \(\mathbf { R }\) in moving the particle from \(A\) to \(B\) is 21 J . Calculate the value of \(a\).
    \section*{PLEASE DO NOT WRITE ON THIS PAGE}
Edexcel FM1 AS 2018 June Q2
9 marks Standard +0.3
2. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{cfa9b998-d57d-4980-9316-1bddeac55b90-04_267_891_346_687} \captionsetup{labelformat=empty} \caption{Figure 1}
\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.
    VILU SIHI NI IIIUM ION OCVGHV SIHILNI IMAM ION OOVJYV SIHI NI JIIYM ION OC
Edexcel FM1 AS 2019 June Q3
7 marks Standard +0.3
  1. A particle, \(P\), of mass \(m \mathrm {~kg}\) is projected with speed \(5 \mathrm {~ms} ^ { - 1 }\) down a line of greatest slope of a rough plane. The plane is inclined to the horizontal at an angle \(\alpha\), where \(\sin \alpha = \frac { 3 } { 5 }\) The total resistance to the motion of \(P\) is a force of magnitude \(\frac { 1 } { 5 } m g\) Use the work-energy principle to find the speed of \(P\) at the instant when it has moved a distance 8 m down the plane from the point of projection.
Edexcel FM1 AS 2020 June Q4
11 marks Standard +0.3
  1. A small ball, of mass \(m\), is thrown vertically upwards with speed \(\sqrt { 8 g H }\) from a point \(O\) on a smooth horizontal floor. The ball moves towards a smooth horizontal ceiling that is a vertical distance \(H\) above \(O\). The coefficient of restitution between the ball and the ceiling is \(\frac { 1 } { 2 }\) In a model of the motion of the ball, it is assumed that the ball, as it moves up or down, is subject to air resistance of constant magnitude \(\frac { 1 } { 2 } \mathrm { mg }\).
    Using this model,
    1. use the work-energy principle to find, in terms of \(g\) and \(H\), the speed of the ball immediately before it strikes the ceiling,
    2. find, in terms of \(g\) and \(H\), the speed of the ball immediately before it strikes the floor at \(O\) for the first time.
    In a simplified model of the motion of the ball, it is assumed that the ball, as it moves up or down, is subject to no air resistance. Using this simplified model,
  2. explain, without any detailed calculation, why the speed of the ball, immediately before it strikes the floor at \(O\) for the first time, would still be less than \(\sqrt { 8 g H }\)
Edexcel FM1 AS 2021 June Q1
7 marks Standard +0.3
1. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{05f6f654-05e5-41d5-a6e4-11cd91a6df83-02_826_700_244_550} \captionsetup{labelformat=empty} \caption{Figure 1}
\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.
Edexcel FM1 AS 2022 June Q3
12 marks Standard +0.3
  1. A plane is inclined to the horizontal at an angle \(\alpha\), where \(\tan \alpha = \frac { 3 } { 4 }\)
A particle \(P\) is held at rest at a point \(A\) on the plane.
The particle \(P\) is then projected with speed \(25 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) from \(A\), up a line of greatest slope of the plane. In an initial model, the plane is modelled as being smooth and air resistance is modelled as being negligible. Using this model and the principle of conservation of mechanical energy,
  1. find the speed of \(P\) at the instant when it has travelled a distance \(\frac { 25 } { 6 } \mathrm {~m}\) up the plane from \(A\). In a refined model, the plane is now modelled as being rough, with the coefficient of friction between \(P\) and the plane being \(\frac { 3 } { 5 }\) Air resistance is still modelled as being negligible.
    Using this refined model and the work-energy principle,
  2. find the speed of \(P\) at the instant when it has travelled a distance \(\frac { 25 } { 6 } \mathrm {~m}\) up the plane from \(A\).
Edexcel FM1 AS 2023 June Q3
10 marks Standard +0.3
  1. A stone of mass 0.5 kg is projected vertically upwards with a speed \(U \mathrm {~ms} ^ { - 1 }\) from a point \(A\). The point \(A\) is 2.5 m above horizontal ground.
The speed of the stone as it hits the ground is \(25 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) The motion of the stone from the instant it is projected from \(A\) until the instant it hits the ground is modelled as that of a particle moving freely under gravity.
  1. Use the model and the principle of conservation of mechanical energy to find the value of \(U\). In reality, the stone will be subject to air resistance as it moves from \(A\) to the ground.
  2. State how this would affect your answer to part (a). The ground is soft and the stone sinks a vertical distance \(d \mathrm {~cm}\) into the ground. The resistive force exerted on the stone by the ground is modelled as a constant force of magnitude 2000 N and the stone is modelled as a particle.
  3. Use the model and the work-energy principle to find the value of \(d\), giving your answer to 3 significant figures.
Edexcel FM2 AS 2022 June Q4
10 marks Standard +0.3
  1. A particle \(P\) moves on the \(x\)-axis. At time \(t\) seconds the velocity of \(P\) is \(v \mathrm {~ms} ^ { - 1 }\) in the direction of \(x\) increasing, where
$$v = \frac { 1 } { 2 } \left( 3 \mathrm { e } ^ { 2 t } - 1 \right) \quad t \geqslant 0$$ The acceleration of \(P\) at time \(t\) seconds is \(a \mathrm {~ms} ^ { - 2 }\)
  1. Show that \(a = 2 v + 1\)
  2. Find the acceleration of \(P\) when \(t = 0\)
  3. Find the exact distance travelled by \(P\) in accelerating from a speed of \(1 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) to a speed of \(4 \mathrm {~m} \mathrm {~s} ^ { - 1 }\)
CAIE M1 2014 November Q6
9 marks Standard +0.3
  1. the work done against the frictional force acting on \(B\),
  2. the loss of potential energy of the system,
  3. the gain in kinetic energy of the system. At the instant when \(B\) has moved 0.9 m the string breaks. \(A\) is at a height of 0.54 m above a horizontal floor at this instant.
    (ii) Find the speed with which \(A\) reaches the floor. \(6 \quad A B C\) is a line of greatest slope of a plane inclined at angle \(\alpha\) to the horizontal, where \(\sin \alpha = 0.28\) and \(\cos \alpha = 0.96\). The point \(A\) is at the top of the plane, the point \(C\) is at the bottom of the plane and the length of \(A C\) is 5 m . The part of the plane above the level of \(B\) is smooth and the part below the level of \(B\) is rough. A particle \(P\) is released from rest at \(A\) and reaches \(C\) with a speed of \(2 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). The coefficient of friction between \(P\) and the part of the plane below \(B\) is 0.5 . Find
    1. the acceleration of \(P\) while moving
      1. from \(A\) to \(B\),
      2. from \(B\) to \(C\),
      3. the distance \(A B\),
      4. the time taken for \(P\) to move from \(A\) to \(C\).
OCR Further Mechanics 2018 March Q1
6 marks Standard +0.3
1 A particle \(P\) of mass 4.2 kg is free to move along the \(x\)-axis which is horizontal. \(P\) is projected from the origin, \(O\), in the positive \(x\) direction with a speed of \(2 \mathrm {~ms} ^ { - 1 }\). As \(P\) moves between \(O\) and the point \(A\) where \(x = 4\), it is acted upon by a variable force of magnitude \(\left( 12 x - 3 x ^ { 2 } \right) \mathrm { N }\) acting in the direction \(O A\).
  1. Calculate the work done by the force as \(P\) moves from \(O\) to \(A\).
  2. Hence, assuming that no other force acts on \(P\), calculate the speed of \(P\) at \(A\).
AQA M2 2010 January Q1
3 marks Easy -1.2
1 An inextensible rope is attached to a sledge which is at rest on a horizontal surface. A constant force of magnitude 40 newtons at an angle of \(30 ^ { \circ }\) to the horizontal is applied to the sledge, as shown in the diagram. \includegraphics[max width=\textwidth, alt={}, center]{06b431ca-d3a8-46d6-b9f8-bac08d3fd51e-2_312_1086_664_466} Calculate the work done by the force as the sledge is moved 5 metres along the surface.
AQA Further AS Paper 2 Mechanics 2018 June Q6
7 marks Standard +0.3
6 At a fairground a dodgem car is moving in a straight horizontal line towards a side wall that is perpendicular to the velocity of the car. The speed of the car is \(1.8 \mathrm {~ms} ^ { - 1 }\) It collides with the side wall and rebounds along its original path with a speed of \(1.2 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) The total mass of the dodgem car and the passengers is 250 kg
6
  1. Find the magnitude of the impulse on the car during the collision with the side wall.
    6
  2. A possible model for the magnitude of the force, \(F\) newtons, acting on the dodgem car due to its collision with the side wall is given by $$F = k t ( 4 - 5 t ) \quad \text { for } 0 \leq t \leq 0.8$$ 6 (b) (i) Find the value of \(k\).
    (b) (ii) Determine the maximum magnitude of the force predicted by the model. 6 (b) (ii) Determine the maximum magnitude of the fored bed bed at
AQA Further AS Paper 2 Mechanics 2022 June Q1
1 marks Easy -1.8
1 A box is being pushed in a straight line along horizontal ground by a force.
The force is applied in the direction of motion and has magnitude 10 newtons. The box moves 5 metres in 2 seconds. Calculate the work done by the force.
Circle your answer.
20 J
25 J
50 J
100 J
AQA Further AS Paper 2 Mechanics 2022 June Q20
1 marks Easy -1.8
20 J
25 J
50 J
100 J 2 Two particles of equal mass are moving on a horizontal surface when they collide.
Immediately before the collision, their velocities are \(\left[ \begin{array} { l } 2 \\ 4 \end{array} \right] \mathrm { ms } ^ { - 1 }\) and \(\left[ \begin{array} { c } 6 \\ - 2 \end{array} \right] \mathrm { ms } ^ { - 1 }\) As a result of the collision the particles coalesce to become a single particle.
Find the velocity of the single particle, immediately after the collision.
Circle your answer.
[0pt] [1 mark] \(\left[ \begin{array} { l } 4 \\ 1 \end{array} \right] \mathrm { m } \mathrm { s } ^ { - 1 }\) \(\left[ \begin{array} { l } 4 \\ 3 \end{array} \right] \mathrm { m } \mathrm { s } ^ { - 1 }\) \(\left[ \begin{array} { l } 8 \\ 2 \end{array} \right] \mathrm { ms } ^ { - 1 }\) \(\left[ \begin{array} { l } 8 \\ 6 \end{array} \right] \mathrm { ms } ^ { - 1 }\) 3 In this question use \(g = 9.8 \mathrm {~m} \mathrm {~s} ^ { - 2 }\) A ball of mass of 0.75 kg is thrown vertically upwards with an initial speed of \(12 \mathrm {~ms} ^ { - 1 }\) The ball is thrown from ground level. 3
  1. Calculate the initial kinetic energy of the ball. 3
  2. The maximum height of the ball above the ground is \(h\) metres.
    Jeff and Gurjas use an energy method to find \(h\) Jeff concludes that \(h = 7.3\) Gurjas concludes that \(h < 7.3\) Explain the reasoning that they have used, showing any calculations that you make.
    4 Wavelength is defined as the distance from the highest point on one wave to the highest point on the next wave. Surfers classify waves into one of several types related to their wavelengths.
    Two of these classifications are deep water waves and shallow water waves.
    4
    1. The wavelength \(w\) of a deep water wave is given by $$w = \frac { g t ^ { 2 } } { k }$$ where \(g\) is the acceleration due to gravity and \(t\) is the time period between consecutive waves. Given that the formula for a deep water wave is dimensionally consistent, show that \(k\) is a dimensionless constant. 4
    2. The wavelength \(w\) of a shallow water wave is given by $$w = ( g d ) ^ { \alpha } t ^ { \beta }$$ where \(g\) is the acceleration due to gravity, \(d\) is the depth of water and \(t\) is the time period between consecutive waves. Use dimensional analysis to find the values of \(\alpha\) and \(\beta\) 5 A car, of mass 1000 kg , is travelling on a straight horizontal road. When the car travels at a speed of \(v \mathrm {~ms} ^ { - 1 }\), it experiences a resistance force of magnitude \(25 v\) newtons. The car has a maximum speed of \(72 \mathrm {~km} \mathrm {~h} ^ { - 1 }\) on the straight road.
      Find the maximum power output of the car.
      Fully justify your answer.
OCR FM1 AS 2021 June Q4
14 marks Standard +0.3
4
2.4
&
B1 for each of two correct statements about the models.
If commenting on the accuracy of (a), must emphasise that (a) is very inaccurate or at least quite inaccurate
Do not allow e.g.
- model (a) is not very effective
- Neither model is accurate
- (a) and (b) are not very accurate
Clear comparison between the accuracy of the two models (must emphasise that (b) is fairly accurate or considerably more accurate than (a)), or other suitable distinct second comment
Do not allow e.g.
- model (b) is more accurate than model (a)
- (b) is not accurate
Do not allow statement claiming that resistance is proportional to speed, or to speed \({ } ^ { 2 }\)
Suitable comments for (a):
- is very inaccurate
- predicted speed is nearly three times the actual value
- constant resistance is not a suitable model
- both models underestimate the resistance (as top speed is lower than expected)
For the linear model (b)
- is fairly accurate (but probably underestimates the resistance at higher speeds)
- resistance is not proportional to speed but is a much better model than constant resistance
3(a)\(T _ { 2 } \cos \theta = m _ { 2 } g\) \(T _ { 2 } = \frac { m _ { 2 } \times 9.8 } { 0.8 } = 12.25 m _ { 2 }\)
M1
A1
[2]
1.1a
1.1
Resolving \(T _ { 2 }\) vertically and balancing forces on \(R\)
Do not allow extra forces present
Allow use of g, e.g. \(\frac { 5 } { 4 } g m _ { 2 }\)
In this solution \(\theta\) is the angle between \(R P\) and \(R A\) Sin may be seen instead if \(\theta\) is measured horizontally.
Do not allow incomplete expressions e.g. \(\frac { m _ { 2 } g } { \sin 53.13 }\)
3(b)(i)\(\begin{aligned}T _ { 2 } \cos \theta + m _ { 1 } g = T _ { 1 } \cos \theta
T _ { 1 } = T _ { 2 } + \frac { 9.8 m _ { 1 } } { 0.8 } =
\qquad 12.25 m _ { 2 } + 12.25 m _ { 1 } = \frac { 49 } { 4 } \left( m _ { 1 } + m _ { 2 } \right) \end{aligned}\)
M1
A1
[2]
3.1b
2.1
Vertical forces on \(P\); 3 terms including resolving of \(T _ { 1 }\); allow sign error
AG Dividing by \(\cos \theta ( = 0.8 )\), substituting their \(T _ { 2 }\) and rearranging
Allow 12.25 instead of \(\frac { 49 } { 4 }\)
Or \(T _ { 1 } \cos \theta = m _ { 1 } g + m _ { 2 } g\) (equation for the system as a whole)
At least one intermediate step must be seen
3(b)(ii)\(\begin{aligned}T _ { 1 } \sin \theta + T _ { 2 } \sin \theta = m _ { 1 } a
12.25 \left( m _ { 1 } + m _ { 2 } \right) \times 0.6 + 12.25 m _ { 2 } \times 0.6 = m _ { 1 } \times 0.6 \omega ^ { 2 }
\omega ^ { 2 } = \frac { 7.35 m _ { 1 } + 14.7 m _ { 2 } } { 0.6 m _ { 1 } } = \frac { 49 \left( m _ { 1 } + 2 m _ { 2 } \right) } { 4 m _ { 1 } } \end{aligned}\)
M1
M1
A1
[3]
3.1b
1.1
2.1
NII horizontally for \(P ; 3\) terms including resolving of tensions; allow sign error
Substituting for \(T _ { 1 }\), their \(T _ { 2 } , \sin \theta\) and \(\alpha\)
AG Must see an intermediate step
Could see \(a\) or \(0.6 \omega ^ { 2 }\) or \(\frac { v ^ { 2 } } { 0.6 }\) or \(\omega ^ { 2 } r\) or \(\frac { v ^ { 2 } } { r } \sin \theta = 0.6\)
must be \(a = 0.6 \omega ^ { 2 }\)
3(c)\(\begin{aligned}\text { E.g } m _ { 1 } \gg m _ { 2 } \Rightarrow \frac { 2 m _ { 2 } } { m _ { 1 } } \approx 0 \text { or } \frac { 49 m _ { 2 } } { 4 m _ { 1 } } \approx 0
\omega \approx \sqrt { \frac { 49 m } { 4 m } } = 3.5 \end{aligned}\)
M1 A1
[2]
1.1
1.1
Allow argument such as if \(m _ { 1 } \gg m _ { 2 }\) then \(m _ { 1 } + 2 m _ { 2 } \approx m _ { 1 }\)
AG \(m\) may be missing
SC1 for result following argument that \(m _ { 2 }\) is negligible (by comparison with \(m _ { 1 }\) ) without justification, or using trial values of \(m _ { 1 }\) and \(m _ { 2 }\) with \(m _ { 1 } \gg m _ { 2 }\).
Do not allow the assumption that \(m _ { 2 } = 0\)
If using trial values, \(m _ { 1 }\) must be at least \(70 \times m _ { 2 }\) to give \(\omega = 3.5\) to 1 dp .
3\multirow{3}{*}{(d)}
\(v = r \omega = 0.6 \sqrt { \frac { 49 \times 2.5 + 98 \times 2.8 } { 4 \times 2.5 } }\)
Final energy \(= 2.5 \times g \times 1\) \(\text { Initial } \mathrm { KE } = \frac { 1 } { 2 } \times 2.5 \times 0.6 ^ { 2 } \times \frac { 49 \times 2.5 + 98 \times 2.8 } { 4 \times 2.5 }\)
Initial PE \(= 2.5 \times g \times 1.2 + 2.8 \times g \times 0.4\)
Energy loss \(= 17.8605 + 40.376 - 24.5 = 33.7365\)
M1
B1
M1
M1
A1
1.2
1.1
1.1
1.1
3.2a
Use of \(v = r \omega\) with values for \(m _ { 1 }\) and \(m _ { 2 }\)
(Assuming zero PE level at 2 m below \(A\); other values possible)
Do not allow use of \(\omega = 3.5\)
oe with different zero PE level awrt 33.7
( \(v = 3.78 , v ^ { 2 } = 14.2884\) )
NB \(\omega = 6.3\) (24.5)
(17.8605)
(40.376)
Alternate method \(v = r \omega = 0.6 \sqrt { \frac { 49 \times 2.5 + 98 \times 2.8 } { 4 \times 2.5 } }\)
Initial KE \(= \frac { 1 } { 2 } \times 2.5 \times 0.6 ^ { 2 } \times \frac { 49 \times 2.5 + 98 \times 2.8 } { 4 \times 2.5 }\)
\(\triangle P E\) for \(m _ { 1 } = \pm 2.5 \times 9.8 \times ( 0.8 - 1 )\)
\(\triangle P E\) for \(m _ { 2 } = \pm 2.8 \times 9.8 ( 1.6 - 2 )\)
Energy loss \(= 17.8605 + 4.9 + 10.976\)
M1
M1
M1
M1
A1
Use of \(v = r \omega\) with values for \(m _ { 1 }\) and \(m _ { 2 }\)
Or \(- \triangle P E\) \(= 2.5 \times 9.8 \times 0.2 + 2.8 \times 9.8 \times 0.4\)
awrt 33.7
( \(v = 3.78 , v ^ { 2 } = 14.2884\) ) \(\mathrm { NB } \omega = 6.3\)
(17.8605)
\(( \pm 4.9 )\)
\(( \pm 10.976 )\)
\(( \pm 15.876 )\)
Or 15.876 + 17.8605
[5]
OCR FM1 AS 2021 June Q2
14 marks Moderate -0.3
2 A particle \(P\) of mass 2.4 kg is moving in a straight line \(O A\) on a horizontal plane. \(P\) is acted on by a force of magnitude 30 N in the direction of motion. The distance \(O A\) is 10 m . \begin{enumerate}[label=(\alph*)] \item Find the work done by this force as \(P\) moves from \(O\) to \(A\). The motion of \(P\) is resisted by a constant force of magnitude \(R \mathrm {~N}\). The velocity of \(P\) increases from \(12 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at \(O\) to \(18 \mathrm {~ms} ^ { - 1 }\) at \(A\). \item Find the value of \(R\). \item Find the average power used in overcoming the resistance force on \(P\) as it moves from \(O\) to \(A\). When \(P\) reaches \(A\) it collides directly with a particle \(Q\) of mass 1.6 kg which was at rest at \(A\) before the collision. The impulse exerted on \(Q\) by \(P\) as a result of the collision is 17.28 Ns . \item
  1. Find the speed of \(Q\) after the collision.
  2. Hence show that the collision is inelastic. It is required to model the motion of a car of mass \(m \mathrm {~kg}\) travelling at a constant speed \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) around a circular portion of banked track. The track is banked at \(30 ^ { \circ }\) (see diagram). \includegraphics[max width=\textwidth, alt={}, center]{b9741472-f230-4e2d-9c8b-47f7e168e938-03_355_565_269_274} In a model, the following modelling assumptions are made.
    For a particular portion of banked track, \(r = 24\).
    (b) Find the value of \(v\) as predicted by the model. A car is being driven on this portion of the track at the constant speed calculated in part (b). The driver finds that in fact he can drive a little slower or a little faster than this while still moving in the same horizontal circle.
    (c) Explain
Edexcel M2 2024 October Q3
Standard +0.3
3. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{3e78f951-041d-4227-aa4b-e67a6ab5b4cd-06_275_1143_303_461} \captionsetup{labelformat=empty} \caption{Figure 1}
\end{figure} A van of mass 900 kg is moving up a straight road inclined at an angle \(\alpha\) to the horizontal, where \(\sin \alpha = \frac { 1 } { 25 }\). The van is towing a trailer of mass 200 kg . The trailer is attached to the van by a rigid towbar which is parallel to the direction of motion of the van and the trailer, as shown in Figure 1. The resistance to the motion of the van from non-gravitational forces is modelled as a constant force of magnitude 400 N .
The resistance to the motion of the trailer from non-gravitational forces is modelled as a constant force of magnitude 240 N . The towbar is modelled as a light rod.
The engine of the van is working at a constant rate of 15 kW .
  1. Find the acceleration of the van at the instant when the speed of the van is \(12 \mathrm {~ms} ^ { - 1 }\) At the instant when the speed of the van is \(14 \mathrm {~ms} ^ { - 1 }\), the trailer is passing the point \(A\) on the slope and the towbar breaks. The trailer continues to move up the slope until it comes to rest at the point \(B\).
    The resistance to the motion of the trailer from non-gravitational forces is still modelled as a constant force of magnitude 240 N .
  2. Use the work-energy principle to find the distance \(A B\).
Pre-U Pre-U 9795/2 2018 June Q7
Moderate -0.3
7 A car has mass 800 kg .
  1. The car accelerates from \(10 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) to \(20 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) in climbing a hill with a vertical height of 16 m . Ignoring resistive forces, find the work done by the engine.
  2. The engine produces a constant power output of 189 kW . The car now travels along horizontal ground. Modelling the resistive force as \(7 v ^ { 2 } \mathrm {~N}\), where \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) is the speed, find the value of \(v\) for which the speed of the car is constant.
CAIE M1 2021 June Q1
3 marks Moderate -0.5
A winch operates by means of a force applied by a rope. The winch is used to pull a load of mass 50 kg up a line of greatest slope of a plane inclined at 60° to the horizontal. The winch pulls the load a distance of 5 m up the plane at constant speed. There is a constant resistance to motion of 100 N. Find the work done by the winch. [3]
CAIE M1 2021 June Q7
11 marks Standard +0.3
\includegraphics{figure_7} A slide in a playground descends at a constant angle of 30° for 2.5 m. It then has a horizontal section in the same vertical plane as the sloping section. A child of mass 35 kg, modelled as a particle \(P\), starts from rest at the top of the slide and slides straight down the sloping section. She then continues along the horizontal section until she comes to rest (see diagram). There is no instantaneous change in speed when the child goes from the sloping section to the horizontal section. The child experiences a resistance force on the horizontal section of the slide, and the work done against the resistance force on the horizontal section of the slide is 250 J per metre.
  1. It is given that the sloping section of the slide is smooth.
    1. Find the speed of the child when she reaches the bottom of the sloping section. [3]
    2. Find the distance that the child travels along the horizontal section of the slide before she comes to rest. [2]
  2. It is given instead that the sloping section of the slide is rough and that the child comes to rest on the slide 1.05 m after she reaches the horizontal section. Find the coefficient of friction between the child and the sloping section of the slide. [6]
CAIE M1 2022 June Q5
9 marks Standard +0.3
Two racing cars \(A\) and \(B\) are at rest alongside each other at a point \(O\) on a straight horizontal test track. The mass of \(A\) is 1200 kg. The engine of \(A\) produces a constant driving force of 4500 N. When \(A\) arrives at a point \(P\) its speed is 25 m s\(^{-1}\). The distance \(OP\) is \(d\) m. The work done against the resistance force experienced by \(A\) between \(O\) and \(P\) is 75 000 J.
  1. Show that \(d = 100\). [3]
Car \(B\) starts off at the same instant as car \(A\). The two cars arrive at \(P\) simultaneously and with the same speed. The engine of \(B\) produces a driving force of 3200 N and the car experiences a constant resistance to motion of 1200 N.
  1. Find the mass of \(B\). [3]
  2. Find the steady speed which \(B\) can maintain when its engine is working at the same rate as it is at \(P\). [3]
CAIE M1 2022 June Q6
8 marks Standard +0.3
A car of mass \(900\text{kg}\) is moving up a hill inclined at \(\sin^{-1} 0.12\) to the horizontal. The initial speed of the car is \(11\text{ms}^{-1}\). After \(12\text{s}\), the car has travelled \(150\text{m}\) up the hill and has speed \(16\text{ms}^{-1}\). The engine of the car is working at a constant rate of \(24\text{kW}\).
  1. Find the work done against the resistive forces during the \(12\text{s}\). [5]
  2. The car then travels along a straight horizontal road. There is a resistance to the motion of the car of \((1520 + 4v)\text{N}\) when the speed of the car is \(v\text{ms}^{-1}\). The car travels at a constant speed with the engine working at a constant rate of \(32\text{kW}\). Find this speed. [3]
CAIE M1 2023 June Q7
11 marks Standard +0.3
A car of mass \(1200\) kg is travelling along a straight horizontal road. The power of the car's engine is constant and is equal to \(16\) kW. There is a constant resistance to motion of magnitude \(500\) N.
  1. Find the acceleration of the car at an instant when its speed is \(20\) m s\(^{-1}\). [3]
  2. Assuming that the power and the resistance forces remain unchanged, find the steady speed at which the car can travel. [2]
The car comes to the bottom of a straight hill of length \(316\) m, inclined at an angle to the horizontal of \(\sin^{-1}(\frac{4}{65})\). The power remains constant at \(16\) kW, but the magnitude of the resistance force is no longer constant and changes such that the work done against the resistance force in ascending the hill is \(128400\) J. The time taken to ascend the hill is \(15\) s.
  1. Given that the car is travelling at a speed of \(20\) m s\(^{-1}\) at the bottom of the hill, find its speed at the top of the hill. [6]
CAIE M1 2024 June Q1
3 marks Standard +0.3
A cyclist and bicycle have a total mass of 72 kg. The cyclist rides along a horizontal road against a total resistance force of 28 N. Find the total work done by the cyclist to increase his speed from \(8\text{ ms}^{-1}\) to \(16\text{ ms}^{-1}\) while travelling a distance of 100 metres. [3]
CAIE M1 2009 June Q2
3 marks Moderate -0.5
\includegraphics{figure_2} A crate \(C\) is pulled at constant speed up a straight inclined path by a constant force of magnitude \(F\) N, acting upwards at an angle of 15° to the path. \(C\) passes through points \(P\) and \(Q\) which are 100 m apart (see diagram). As \(C\) travels from \(P\) to \(Q\) the work done against the resistance to \(C\)'s motion is 900 J, and the gain in \(C\)'s potential energy is 2100 J. Write down the work done by the pulling force as \(C\) travels from \(P\) to \(Q\), and hence find the value of \(F\). [3]