Video Transcript
An average force of 6250 newtons is
applied by a car’s wheels while the car moves a distance of 880 meters. The car’s engine does 10 megajoules
of work to move the car. How much work in megajoules was
done that did not involve moving the car, such as moving the air around the car,
moving parts of the road surface, and heating parts of the car?
Okay, so in this question, we have
a car. And we’re told that this car moves
a distance of 880 meters and that an average force of 6250 newtons is applied by its
wheels. So, the car is moving in this
direction, and the distance in this direction that it moves is 880 meters, which
we’ve labeled as 𝑑. It’s probably worth saying that if
this diagram was drawn to scale, then that would make this car a few 100 meters
wide. So, clearly, this diagram is not to
scale. Let’s also label the average force
applied by the car’s wheels as 𝐹.
We’re told in the question that the
total amount of work done by the car’s engine is 10 megajoules. But the question also signals to us
that not all of this work actually goes towards moving the car. We can calculate how much work is
done that does involve moving the car using the force 𝐹 that’s applied by the car’s
wheels and the distance 𝑑 that the car moves. We can recall that the work done by
a force to move an object is equal to the magnitude of the force multiplied by the
distance the object moves. So, if we sub our values for the
force 𝐹 and the distance 𝑑 into this equation, then the work that we’ll calculate
is the work that’s done that causes the car to move. Let’s call this work done 𝑊
subscript m, where the m stands for move.
Then, subbing in the values for 𝐹
and 𝑑, we find that 𝑊 subscript m is equal to 6250 newtons multiplied by 880
meters. The force here is expressed in the
SI unit for force, the newton, and the distance is in the SI unit of meters. A force measured in newtons and a
distance in meters will give us a value for the work done in the SI unit of energy,
the joule. When we evaluate the expression, we
get a result of 5.5 times 10 to the sixth joules. This value is the amount of work
that’s done that does go towards moving the car. What the question is asking us to
find is how much work is done that does not involve moving the car.
The other bit of information that
we were given is that the total amount of work done by the engine was 10
megajoules. Let’s label this total work done as
𝑊 subscript t. Now, the total work that’s done by
the engine must be equal to the work done that does move the car plus the work done
that doesn’t go towards moving the car. As we’re told in the question, this
work that doesn’t involve moving the car could be things such as moving the air
around the car, moving parts of the road surface, or heating parts of the car. Let’s label this as 𝑊 subscript
o, where the o stands for other because this is all of the work done that goes to
other things than moving the car.
We’ve already labeled the total
work done as 𝑊 subscript t and the work done that does move the car as 𝑊
subscript m. So, we could rewrite this equation
in terms of symbols as 𝑊 subscript t is equal to 𝑊 subscript m plus 𝑊 subscript
o. The quantity that we’re trying to
find is 𝑊 subscript o, the work done that does not involve moving the car. So, let’s rearrange this equation
to make 𝑊 subscript o the subject. To do this, we subtract 𝑊
subscript m from both sides of the equation. The plus and minus 𝑊 subscript m
terms on the right-hand side then cancel each other out. From here, we can then write this
equation the other way around to say that 𝑊 subscript o is equal to 𝑊 subscript
t minus 𝑊 subscript m.
At the moment, our value for the
total work done by the engine 𝑊 subscript t is given in units of megajoules. Meanwhile, the value we found for
the work done to move the car, 𝑊 subscript m, is in units of joules. Before we sub those values into
this equation, we need them both to be in the same units. Since the question asks us to give
our answer in units of megajoules, then it makes sense for us to convert our value
of 𝑊 subscript m from joules into megajoules.
To do this, we can recall that one
megajoule is equal to 10 to the sixth joules. So, to convert from joules into
megajoules, we take the value in units of joules and we multiply by one over 10 to
the sixth megajoules per joule. In terms of the units, the joules
cancel with the per joule, and this leaves us with units of megajoules. When we evaluate the expression, we
find that 𝑊 subscript m is equal to 5.5 megajoules.
Now that we have values for both 𝑊
subscript t and 𝑊 subscript m in the same units of megajoules, we are ready to go
ahead and sub those values into this equation to calculate the value of 𝑊 subscript
o. When we do the substitution, we
find that 𝑊 subscript o is equal to 10 megajoules minus 5.5 megajoules. Evaluating the expression gives a
result of 4.5 megajoules.
And so, we have found that the
amount of work that was done that did not involve moving the car was 4.5
megajoules.
