Final answer:
The force required to lift 250 kg using a second-class lever system with a resistance arm of 0.1m and a force arm of 0.25m is calculated to be 981 N using the principle of lever equilibrium. None of the options provided match this calculated force.
Step-by-step explanation:
To determine the force required to lift a 250 kg weight using a second-class lever, we have to apply the principle of lever equilibrium. The principle states that in a state of equilibrium, the force multiplied by the force arm equals the resistance multiplied by the resistance arm (Force × Force Arm = Resistance × Resistance Arm).
In the student's calf raise scenario, we are given a resistance (weight to be lifted) of 250 kg, a resistance arm (distance from the fulcrum to the weight's line of action) of 0.1 m, and a force arm (distance from the fulcrum to the point where force is applied) of 0.25 m. First, we convert the resistance (weight) to newtons because the weight is in kilograms. The weight (W) in newtons is calculated using the weight (in kg) multiplied by the acceleration due to gravity (g = 9.81 m/s²): W = mass × g = 250 kg × 9.81 m/s² = 2452.5 N.
Using the lever equilibrium principle, we set up the equation:
Force × 0.25 m = 2452.5 N × 0.1 m
Solving for the force (F) gives us:
F = (2452.5 N × 0.1 m) / 0.25 m
F = 245.25 N / 0.25
F = 981 N
However, none of the options provided match the calculated force of 981 N. Therefore, there may have been an error in the question, the options provided, or further details might be needed for an accurate answer to this specific scenario.