The power expended by the construction hoist to lift the object vertically for 10 seconds is 5000 watts.
To determine the power expended by the construction hoist, we can use the formula for power, which is defined as the rate of doing work. The work done (W) is calculated as the product of force (F) and displacement (d) in the direction of the force, expressed as W=F⋅d.
In this scenario, the construction hoist exerts an upward force (F) of 500 Newtons on an object with a mass (m) of 50 kilograms. The force exerted is against gravity, and the displacement (d) is vertical, considering the object is being lifted vertically.
The work done in lifting the object for a distance d can be determined using the gravitational potential energy formula: W=mgh, where m is the mass, g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the vertical height.
Since the hoist starts from rest, the initial velocity is zero, and we can use the kinematic equation
h= 1/2 gt^2 to express the vertical height in terms of time (t).s
h= 1/2 ⋅9.8⋅(10)^2 =490meters
Now, substitute the values into the work formula:
W=50⋅9.8⋅490=240100Joules
Finally, calculate power (P) using the formula P= W/t, where t is the time duration of 10 seconds: P= 240100/10=24010Watts
Therefore, the power expended by the construction hoist to lift the object vertically for 10 seconds is 24010 Watts, or 24.01 kilowatts.