The power is 833.3 W
Explanation:
First of all, we need to calculate the work done in lifting the barbell, which is equal to the change in gravitational potential energy of the barbell:
where
mg = 1250 N is the weight of the barbell
h = 2 m is the change in height
Substituting,
Now we can calculate the power, which is equal to the work done per unit time:
where
W = 2500 J is the work done
t = 3 s is the time taken
Substituting,
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Answer:
1) The mass of the student is approximately 73.39 kg
2) The net force on the student is approximately 947.523 N
3) The value the scale will read is approximately 96.59 kg
Explanation:
The given parameters are;
The weight of the student = 720 N
The speed at which the elevator is decreasing = 3.1 m/s²
1) The weight of the student = The mass of the student × The acceleration due to gravity
The acceleration due to gravity is a constant = 9.81 m/s²
Substituting the known values gives;
720 N = The mass of the student × 9.81 m/s²
∴ The mass of the student = 720 N/(9.81 m/s²) ≈ 73.39 kg
2) The forces acting on the student are;
i) The force of gravity which is the weight of the student acting downwards
ii) The inertia force of the slowing elevator acting downwards in the same direction as the weight of the student
The net force, = The weight of the student + The inertia force of the slowing elevator
∴ The net force, = 720 N + 73.39 kg × 3.1 m/s² ≈ 947.523 N
3) The scale will read the mass of the student as follows;
Mass reading of student on the scale = Force on scale/9.81
∴ Mass reading of student on the scale = 947.523/9.81 ≈ 96.59 kg
The value the scale will read = 96.59 kg.