| Exam Board | OCR MEI |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2012 |
| Session | January |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Work-energy over time interval |
| Difficulty | Moderate -0.3 This is a straightforward multi-part work-energy question requiring standard formula application (P=W/t, KE=½mv², work-energy principle, P=Fv, resolving forces on slopes). All parts follow directly from bookwork with clear numerical substitution and no conceptual challenges or novel problem-solving required. Slightly easier than average due to its routine nature. |
| Spec | 6.02a Work done: concept and definition6.02b Calculate work: constant force, resolved component6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae6.02l Power and velocity: P = Fv |
Question 1:
A thin rigid non-uniform beam AB of length 6 m has weight 800 N. Its centre of mass, G, is 2 m from B.
Initially the beam is horizontal and in equilibrium when supported by a small round peg at C, 1 m from A, and a light vertical wire at B.
(i) Calculate the tension in the wire and the normal reaction of the peg on the beam. [4]
M1: Takes moments about C (or another point)
A1: Correct tension in wire
A1: Correct normal reaction of peg
The beam is now held horizontal and in equilibrium with the wire at $70°$ to the horizontal. The peg at C is rough and still supports the beam 1 m from A. The beam is on the point of slipping.
(ii) Calculate the new tension in the wire. Calculate also the coefficient of friction between the peg and the beam. [7]
M1: Resolves forces horizontally
M1: Resolves forces vertically
M1: Takes moments about C
A1: Correct new tension in wire
A1: Correct friction force
A1: Correct coefficient of friction
The beam is now held in equilibrium at $30°$ to the vertical with the wire at $\theta°$ to the beam. A new small smooth peg now makes contact with the beam at C, still 1 m from A. The tension in the wire is now T N.
(iii) By taking moments about C, resolving in a suitable direction and obtaining two equations in terms of $\theta$ and T, or otherwise, calculate $\theta$ and T. [7]
M1: Takes moments about C
M1: Resolves in suitable direction
A1: Obtains correct equation in $\theta$ and T (first equation)
A1: Obtains correct equation in $\theta$ and T (second equation)
A1: Correct value of $\theta$
A1: Correct value of T1 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.\\
(i) 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.\\
(ii) Calculate the change in the kinetic energy of the bus.\\
(iii) 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.\\
(iv) 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.\\
(v) Calculate the power of the driving force on the bus.
\hfill \mbox{\textit{OCR MEI M2 2012 Q1 [17]}}