Step-by-step Solution:

1. Identify the given equation and known values. The energy of the system is given as a function of time t: $E(t) = A^2 \exp(-\alpha t)$ The given values are:

   • Constant, $\alpha = 0.2\,s^{-1}$
   • Percentage error in measurement of A: (%ΔA) = 1.25%
   • Percentage error in measurement of time t: (%Δt) = 1.50%
   • The specific time at which the error needs to be calculated: t = 5 s

2. Use logarithmic differentiation to find the relation for fractional errors. To find the relative error in E, we can take the natural logarithm of both sides of the energy equation: $\ln(E) = \ln(A^2 \exp(-\alpha t))$ Using the properties of logarithms, $\ln(ab) = \ln(a) + \ln(b)$ and $\ln(x^n) = n\ln(x)$: $\ln(E) = \ln(A^2) + \ln(\exp(-\alpha t))$ $\ln(E) = 2\ln(A) - \alpha t$

3. Differentiate the logarithmic expression to find the relationship between small changes. Differentiating the above expression with respect to the variables A and t (treating α as a constant with no error), we get: $\frac{dE}{E} = 2 \frac{dA}{A} - \alpha \, dt$

4. Formulate the expression for the maximum percentage error. For small finite changes, we can approximate differentials with deltas (Δ). The maximum possible error is found by adding the absolute values of the individual error terms: $\left| \frac{\Delta E}{E} \right|_{\text{max}} = \left| 2 \frac{\Delta A}{A} \right| + \left| -\alpha \, \Delta t \right|$ $\frac{\Delta E}{E} = 2 \frac{\Delta A}{A} + \alpha \, \Delta t$ We are given the percentage error in t, which is $\frac{\Delta t}{t} \times 100$. So we can write $\Delta t = t \left( \frac{\Delta t}{t} \right)$. Substituting this into the equation: $\frac{\Delta E}{E} = 2 \frac{\Delta A}{A} + \alpha t \left( \frac{\Delta t}{t} \right)$ To get the percentage error, we multiply the entire equation by 100: $\left( \frac{\Delta E}{E} \times 100 \right) = 2 \left( \frac{\Delta A}{A} \times 100 \right) + \alpha t \left( \frac{\Delta t}{t} \times 100 \right)$ This can be written concisely as: $\% \Delta E = 2 (\% \Delta A) + (\alpha t) (\% \Delta t)$

5. Substitute the given numerical values to calculate the final answer. First, calculate the dimensionless product αt: $\alpha t = (0.2\,s^{-1}) \times (5\,s) = 1$ Now, substitute this and the given percentage errors into the formula: $\% \Delta E = 2 (1.25\%) + (1) (1.50\%)$ $\% \Delta E = 2.5\% + 1.5\%$ $\% \Delta E = 4.0\%$

6. Final Answer. The percentage error in the value of E(t) at t = 5 s is 4.