How to answer numerical questions
Here is quite a complicated question which illustrates the principles of problem solving.
A lorry of mass 3000 kg, whose registration number is HG05ABC, starts from rest when the traffic lights go green, and travels for half a minute, mowing down any pedestrians in its path, with an acceleration of 1 m/s² until it reaches a motorway intersection. During this time there is a frictional force of 1.2 kN acting on the vehicle. Calculate the work done by the engine during the journey from the traffic lights to the motorway.
This looks hard(ish) because the data you need in order to answer the final question is apparently not present. It has to be unpacked first. (For the Questions A page, which has easier questions than this, you only need the three asterisked paragraphs of what follows.)
*Underline (or highlight) any portions of the question that give you numerical data either directly or implied.
Ignore clutter that is just dressing the question up.
A lorry of mass 3000 kg, whose registration number is HG05ABC, starts from rest when the traffic lights go green, and travels for half a minute, mowing down any pedestrians in its path, with an acceleration of 1 m/s² until it reaches a motorway intersection. During this time there is a frictional force of 1.2 kN acting on the vehicle.
*Translate these highlighted portions into data statements.
Use the standard letter and an equal sign, and convert the units into SI. Translate the multipliers
m = 3000 kg  Note that kg is already SI. It is the one exception to the "translate the multipliers" rule. 
u = 0 m/s  'rest' means a zero velocity. u is the letter for initial velocity. 
t = 30 s  The SI unit for time is the second. 1 minute = 60 s 
a = 1 m/s²  
F_{friction} =  1.2 x 10^{3} N 
A subscript has been used because other forces might be involved. A minus is used because the frictional force is opposite to the direction of travel. The multiplier k is translated literally as x 10^{3 } 
Identify any equations from your repertoire that involve the variables you have identified
They should not contain more than one 'extra' variable (i.e. that is not in your list)
s = u t + ½ a t²  These are two of the socalled SUVAT
equations. Each contains just four of s, u, v,
a, t. Given any three of the quantities,
you can always find an equation that contains those three that will
therefore enable you to calculate the other quantity in that equation. The other equations, not allowed here because they contain two 'extra' variables, are: v² = u ² + 2 a s s = ½ (u + v) t s = v t  ½ a t² 
v = u + a t  
F = m a  F is the resultant force acting, taking all the individual forces (and their directions) into account 
*Calculate new data using these equations. Here's how:
v = u + a t v = 0 m/s + (1 m/s² x 30 s) v = 30 m/s 
s = u t + ½
a t² s = ½ x 1 m/s² x (30 s)² s = 450 m 
F_{resultant}
= m a F_{resultant} = 3000 kg x 1 m/s² F_{resultant} = 3000 N 
Use this button if your purpose in visiting this page was to answer questions on the Questions A page
Identify any additional equations that could now be used, in the light of these extra data
E_{kinetic} = ½ m v²  These should have the same value, since it is the resultant force that accelerates the lorry, thus enabling it to acquire kinetic energy 
W_{resultant} = F_{resultant} ´ s  
W_{friction} = F_{friction} ´ s  Although F_{friction} is negative, W_{friction} has to be accorded a positive value in the end. 
F_{resultant} = F_{engine} + F_{friction }  But remember that F_{friction} is negative so a subtraction will eventually be carried out. 
Calculate any additional data that are now accessible.
E_{kinetic}
= ½ m v² E_{kinetic} = ½ ´ 3000 kg ´ (30 m/s)² E_{kinetic} = 1.35 ´ 10^{6 }J

W_{friction}
= F_{friction}
´ s W_{friction} = 1.2 ´ 10^{3} N ´ 450 m W_{friction} = 5.4 ´ 10^{5 }J

F_{resultant}
= F_{engine}
+ F_{friction
}
F_{engine} = F_{resultant } F_{friction} F_{engine} = 3000 N  (1200 N) F_{engine} = 4200 N 
Continue this process of identifying new equations and generating new data for so long as is necessary (or possible)
In this particular problem, we can now proceed to a solution by noticing that the total energy delivered by the engine is used either to give the lorry kinetic energy or to work against friction. Thus
E_{engine} = E_{kinetic} + W_{friction}
E_{engine} = 1.35 ´ 10^{6 }J + 5.4 ´ 10^{5 }J
E_{engine} = 1.89 ´ 10^{6 }J 
If we had not noticed this, the next cycle of equations would have yielded
W_{engine} = F_{engine} ´ s
W_{engine} = 4200 N ´ 450 m
W_{engine} = 1.89 ´ 10^{6 }J 
Finally, of course, we must respect the fact that the acceleration datum is given to just one significant figure. So our final answer is
W_{engine} = 2 MJ 