Answer :
To calculate the mass that Sarita lifted given the power, height, and time, we can use the relationship between power, work, and energy. Let’s break it down step by step:
1. Power (P) is the rate at which work is done. It is given by the formula:
[tex]\[ P = \frac{\text{Work}}{\text{time}} \][/tex]
Here, the power provided is [tex]\(480 \, \text{watts}\)[/tex].
2. Work (W) done is equivalent to the gravitational potential energy gained by the mass lifted. This can be calculated using:
[tex]\[ W = m \cdot g \cdot h \][/tex]
where [tex]\(m\)[/tex] is the mass, [tex]\(g\)[/tex] is the acceleration due to gravity ([tex]\(9.81 \, \text{m/s}^2\)[/tex]), and [tex]\(h\)[/tex] is the height ([tex]\(10 \, \text{meters}\)[/tex]).
3. Rearrange the power formula to solve for the work done:
[tex]\[ \text{Work} = P \cdot \text{time} \][/tex]
Given the power ([tex]\(480 \, \text{watts}\)[/tex]) and the time ([tex]\(12 \, \text{seconds}\)[/tex]):
[tex]\[ \text{Work} = 480 \, \text{watts} \cdot 12 \, \text{seconds} = 5760 \, \text{joules} \][/tex]
4. Substitute this value of work into the potential energy formula to solve for [tex]\(m\)[/tex]:
[tex]\[ 5760 \, \text{joules} = m \cdot 9.81 \, \text{m/s}^2 \cdot 10 \, \text{meters} \][/tex]
5. Isolate [tex]\(m\)[/tex] (mass) to find its value:
[tex]\[ m = \frac{5760 \, \text{joules}}{9.81 \, \text{m/s}^2 \cdot 10 \, \text{meters}} \][/tex]
[tex]\[ m \approx 58.716 \, \text{kg} \][/tex]
Conclusion:
So, Sarita lifted approximately [tex]\(58.716 \, \text{kg}\)[/tex] up to [tex]\(10 \, \text{meters}\)[/tex] in [tex]\(12 \, \text{seconds}\)[/tex].
1. Power (P) is the rate at which work is done. It is given by the formula:
[tex]\[ P = \frac{\text{Work}}{\text{time}} \][/tex]
Here, the power provided is [tex]\(480 \, \text{watts}\)[/tex].
2. Work (W) done is equivalent to the gravitational potential energy gained by the mass lifted. This can be calculated using:
[tex]\[ W = m \cdot g \cdot h \][/tex]
where [tex]\(m\)[/tex] is the mass, [tex]\(g\)[/tex] is the acceleration due to gravity ([tex]\(9.81 \, \text{m/s}^2\)[/tex]), and [tex]\(h\)[/tex] is the height ([tex]\(10 \, \text{meters}\)[/tex]).
3. Rearrange the power formula to solve for the work done:
[tex]\[ \text{Work} = P \cdot \text{time} \][/tex]
Given the power ([tex]\(480 \, \text{watts}\)[/tex]) and the time ([tex]\(12 \, \text{seconds}\)[/tex]):
[tex]\[ \text{Work} = 480 \, \text{watts} \cdot 12 \, \text{seconds} = 5760 \, \text{joules} \][/tex]
4. Substitute this value of work into the potential energy formula to solve for [tex]\(m\)[/tex]:
[tex]\[ 5760 \, \text{joules} = m \cdot 9.81 \, \text{m/s}^2 \cdot 10 \, \text{meters} \][/tex]
5. Isolate [tex]\(m\)[/tex] (mass) to find its value:
[tex]\[ m = \frac{5760 \, \text{joules}}{9.81 \, \text{m/s}^2 \cdot 10 \, \text{meters}} \][/tex]
[tex]\[ m \approx 58.716 \, \text{kg} \][/tex]
Conclusion:
So, Sarita lifted approximately [tex]\(58.716 \, \text{kg}\)[/tex] up to [tex]\(10 \, \text{meters}\)[/tex] in [tex]\(12 \, \text{seconds}\)[/tex].
