A student wants to use a spring-based scale to find the mass of a delivery parcel. The scale is marked in kilograms instead of Newtons.
The scale is not calibrated correctly: with nothing hanging from it, the scale shows 1.2 kg.
Calculate the weight (in Newtons) of a 1.2 kg mass on Earth, where the gravitational field strength is g=10 N/kgg = 10\text{ N/kg}g=10 N/kg.
The student decides to check the scale. She does not have standard masses, so she puts ceramic mugs into a light plastic carrier bag and hangs it from the scale. Each mug is labeled as having a mass of 0.3 kg.
Her readings are shown in the table:
| Number of mugs | Scale reading (kg) |
|---|---|
| 0 | 1.2 |
| 1 | 1.5 |
| 2 | 1.8 |
| 3 | 2.1 |
| 4 | 2.9 |
| 5 | 2.7 |
| 6 | 3.0 |
Identify the anomalous data point in the table. Briefly explain why it is anomalous and how you would deal with it when plotting a line of best fit.
State the relationship between the number of mugs and the scale reading (excluding the anomalous point).
The student remembers that six mugs gave a scale reading of 3.0 kg. She calculates:
mass of six mugs=6×0.3 kg=1.8 kg \text{mass of six mugs} = 6 \times 0.3\text{ kg} = 1.8\text{ kg} mass of six mugs=6×0.3 kg=1.8 kgand notes that:
3.0 kg−1.2 kg=1.8 kg 3.0\text{ kg} - 1.2\text{ kg} = 1.8\text{ kg} 3.0 kg−1.2 kg=1.8 kgShe concludes: "I can use this scale as normal! All I need to do is subtract 1.2 kg from each reading to get the correct mass."
She hangs her delivery parcel from the scale. The scale reading is 4.5 kg. She concludes that her parcel must have a mass of exactly 3.3 kg.
Suggest three reasons why the student's conclusions might be incorrect.
