2019 A/L Physics Past Paper · Question 14 (English Medium)

Answer: (5) $\frac{1}{2}(T_2 + T_1)(l_2 - l_1)$

What the question asks

We want:

The energy stored in the wire when the tension goes slowly from $T_1$ to $T_2$.

The length goes from $l_1$ to $l_2$.

Relevant theory

We pull slowly. So the wire has no kinetic energy.

All the work done is stored in the wire.

Within the proportional limit, Hooke's law holds:

$$T = k(l - l_0)$$

So the graph of $T$ against $l$ is a straight line.

The work done is the area under the graph.

The energy stored in an extended wire is equal to the area of the graph of extension against the load.

Required equations

$T = k\,(l - l_0)$

$W = \int_{l_1}^{l_2} T\,dl$

$W = \frac{1}{2}(T_1 + T_2)(l_2 - l_1)$

Reasoning and calculation

The letters:

$T_1$, $T_2$: the tension at the start and at the end.

$l_1$, $l_2$: the length at the start and at the end.

$l_2 - l_1$: how much the wire stretches.

The tension changes in a straight line.

So we can use its mean value:

$$\dfrac{T_1 + T_2}{2}$$

The mean force times the stretch gives the work.

$W = \frac{T_1 + T_2}{2} \times (l_2 - l_1)$

$W = T_1(l_2 - l_1) + \frac{1}{2}(T_2 - T_1)(l_2 - l_1)$

$W = \frac{1}{2}(T_2 + T_1)(l_2 - l_1)$

Reading the options

Look at the answer.

It is half of the sum of the tensions, times the difference of the lengths.

Only option (5) has this form.

Checks

Check the unit.

$$[T][l] = MLT^{-2} \cdot L = ML^2T^{-2}$$

That is the dimension of energy ✓.

Options (2) and (4) have the same dimension. So this check alone is not enough.

Does it make sense?

Start with no tension, $T_1 = 0$.

Option (5) gives $\dfrac{1}{2}T_2 (l_2 - l_1)$.

That is the $\dfrac{1}{2}Fe$ we know ✓.

If nothing changes, $l_2 = l_1$ and the energy is zero.

Option (5) gives zero ✓. Option (4) does not.

$0.5 * (T1 + T2) * (T2 - T1) / k / 1000$

Conclusion

$W = \frac{1}{2}(T_2 + T_1)(l_2 - l_1)$

The area under the line is the work:

$$W = \dfrac{1}{2}(T_2 + T_1)(l_2 - l_1)$$

So the answer is option (5).

Using $\dfrac{1}{2}Fe$ straight away and getting $\dfrac{1}{2}(T_2 - T_1)(l_2 - l_1)$.

$\dfrac{1}{2}Fe$ is only for a start with no tension. Here the wire already has $T_1$. So add the rectangle too.