How long will it take a 7.08 hp motor to lift a 250 kg beam directly upward at constant velocity from the ground to a height of 45.0 m? Assume frictional forces are negligible.(1 hp = 746 W) A)20.9 s B)1.56 × 10⁴ s C)2.18 × 10⁴ s D)39.7 s

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To find the time, we first need to calculate the power required to lift the beam and then use the power of the motor to find the time. Power (P) is defined as work done (W) per unit time (t), and work done is equal to the force applied (F) times the distance (d). Since the beam is lifted at a constant velocity, the force applied by the motor is equal to the weight of the beam (mass (m) * acceleration due to gravity (g)), and the distance is the height (h) to which the beam is lifted.$$ P = \frac{W}{t} = \frac{F \cdot d}{t} = \frac{m \cdot g \cdot h}{t} $$Given:- Mass of the beam, $m = 250 \, \text{kg}$- Height, $h = 45.0 \, \text{m}$- Acceleration due to gravity, $g = 9.81 \, \text{m/s}^2$- Power of the motor, $P = 7.08 \, \text{hp} = 7.08 \times 746 \, \text{W}$ (since $1 \, \text{hp} = 746 \, \text{W}$)First, convert horsepower to watts:$$ P = 7.08 \times 746 \, \text{W} = 5284.88 \, \text{W} $$Then, solve for time $t$:$$ t = \frac{m \cdot g \cdot h}{P} = \frac{250 \cdot 9.81 \cdot 45.0}{5284.88} \approx 20.9 \, \text{s} $$Therefore, it will take approximately 20.9 seconds for the motor to lift the beam to the height of 45.0 meters.

How Long Will It Take a 7