The amount of work done by the student is 245 Joules. Option B.
The work done, W, for any object is calculated using the formula:
![\[ W = m \cdot g \cdot h \]](https://img.qammunity.org/2024/formulas/physics/high-school/pr8gm991ywtzor3uvzpgy694931cwcrw4d.png)
where:
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is the mass (2.50 kg),
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is the acceleration due to gravity (approximately 9.8 m/s²),
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is the height (10.0 m).
In this case:
![\[ W = 2.50 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 \cdot 10.0 \, \text{m} \]](https://img.qammunity.org/2024/formulas/physics/high-school/mveckgkekmqs19ruub8e1didtqwhyn7655.png)
![\[ W = 245 \, \text{Joules} \]](https://img.qammunity.org/2024/formulas/physics/high-school/x83a8d3dzl3iz8ievy43zhwjo5937qu532.png)
Therefore, the amount of work done by the student is 245 Joules.
The complete question:
A student doing an activity lowers a 2.50 kg mass 10.0 m from the top of stadium bleachers. Which represents the amount of work done by the student
a) 0 J
b) 245 J
c) 2500 J
d) 25,000 J