Calculate the uncertainty in g using the provided slope and its uncertainty.

Question

ScanSolve AI Tutor · Accepted Answer

✅ Correct.

The calculations and the application of the uncertainty propagation formula are correct.

Calculate $S_s$ .
$S_s = \frac{4 \times 0.004}{6 \times 3.90} = \frac{0.016}{23.4} \approx 0.00068376 \text{ m/s}^2$
Rounding to one significant figure, $S_s \approx 0.0007 \text{ m/s}^2$ . This matches the student's calculation.
Use the uncertainty propagation formula.
The formula for propagating relative uncertainty is given as:
$\frac{S_g}{g} = \frac{S_s}{S}$
Rearranging to solve for $S_g$ :
$S_g = \frac{S_s \times g}{S}$
Given:
$S_s = 0.0007 \text{ m/s}^2$ (uncertainty in slope)
$g = 9.8 \text{ m/s}^2$ (assumed value of acceleration due to gravity for calculation of uncertainty)
$S = 0.25 \text{ m/s}^2$ (nominal value of the slope)
Substitute the values and calculate $S_g$ .
$S_g = \frac{0.0007 \text{ m/s}^2 \times 9.8 \text{ m/s}^2}{0.25 \text{ m/s}^2}$ $S_g = \frac{0.00686}{0.25} \text{ m/s}^2$ $S_g = 0.02744 \text{ m/s}^2$
Rounding to two significant figures (consistent with the precision of 9.8 and 0.25, and a common practice for uncertainties), $S_g \approx 0.03 \text{ m/s}^2$ . This matches the student's calculation.
State the final result.
The final result for $g$ is given as $9.86 \pm 0.03 \text{ m/s}^2$ . The calculated uncertainty $0.03 \text{ m/s}^2$ is correctly applied to the experimental value of $g$ (assumed to be $9.86 \text{ m/s}^2$ ). The number of decimal places for the uncertainty and the central value are consistent.

Step-by-step solution