91Ó°ÊÓ

Gravitational force is the force that attracts any object with mass towards the center of the Earth. The force can be calculated using the formula: \ F = m \times g \ where:

• \( m \) is the mass of the object
• \( g \) is the acceleration due to gravity (approximately \( 9.8 \) m/s² on the surface of the Earth)

In this exercise, we calculate the gravitational force for both the elevator cab and its counterweight. For the elevator cab with a mass of \( 1200 \text{ kg} \), the gravitational force is: \( F_{\text{cab}} = 1200 \text{ kg} \times 9.8 \text{ m/s}^2 = 11760 \text{ N} \). Similarly, for the counterweight weighing \( 950 \text{ kg} \), the gravitational force is: \( F_{\text{counterweight}} = 950 \text{ kg} \times 9.8 \text{ m/s}^2 = 9310 \text{ N} \).
Understanding gravitational force is crucial in physics problems involving vertical motion since it directly impacts the net force required to move an object.

In physics, work is done when a force is applied to an object, and the object moves in the direction of the force. The formula to calculate work is: \ W = F \times d \ where:

• \( W \) is the work done (in joules)
• \( F \) is the net force applied (in newtons)
• \( d \) is the distance traveled by the object in the direction of the force (in meters)

In our exercise, we computed the work done by the motor. With a net force of \( 2450 \text{ N} \) and a distance of \( 54 \text{ m} \): \( W_{\text{motor}} = 2450 \text{ N} \times 54 \text{ m} = 132300 \text{ J} \). This amount of work moves the elevator cab upwards against gravitational forces.
Energy is closely related to work, where work done on an object results in a change of its energy. Whether lifting or accelerating an object, work-energy principles are foundational in solving physics problems.

Power in physics is the rate at which work is done. It tells us how much energy is transferred or converted per unit time. Power can be calculated using the formula: \ P = \frac{W}{t} \ where:

• \( P \) is the power (in watts)
• \( W \) is the work done (in joules)
• \( t \) is the time (in seconds)

In this problem, we are asked to find the average power required by the motor. After calculating the work to be \( 132300 \text{ J} \) and converting the time to \( 180 \text{ s} \), the power is: \( P_{\text{average}} = \frac{132300 \text{ J}}{180 \text{ s}} = 735 \text{ W} \).
Understanding power is vital in applications where efficiency and energy transfer over time are critical, such as motors in machines.

Elevator mechanics involves understanding the forces and motions that elevate the cab. In this problem, the elevator has a cab and a counterweight. The motor helps to pull the cab upward due to the insufficient counterweight.
A key aspect here is the balance of forces. The force exerted by the motor must overcome the difference in gravitational forces between the cab and the counterweight. By calculating the net force required (\( 2450 \text{ N} \)), we ensure the cab moves to the desired height smoothly.
Elevators often use counterweights to help balance the load, which makes the motor's job easier and more efficient. This principle of mechanics is fundamental in designing and operating vertical transportation systems like elevators.

Computing the net force is crucial in understanding the actual effort needed to move an object. Net force is the overall force acting on an object when all the individual forces are taken into account. The formula to calculate net force when two opposing forces are present is: \ F_{\text{net}} = F_{\text{applied}} - F_{\text{opposing}} \ . In our elevator exercise, the net force is the difference between the gravitational force on the cab (\( 11760 \text{ N} \)) and the gravitational force on the counterweight (\( 9310 \text{ N} \)): \( F_{\text{net}} = 11760 \text{ N} - 9310 \text{ N} = 2450 \text{ N} \)
This net force helps us understand the required motor force to lift the elevator. Accurately computing net force ensures correct application of subsequent calculations related to work, energy, and power.