How Long Will it Take a 2750-Watt Motor to Lift a 385-kg Piano to a Sixth-Story Window 16.0 Meters Above?

In the realm of physics and engineering, understanding the concept of work, power, and energy conversion plays a crucial role in various practical scenarios. One such scenario involves calculating the time it takes for a specific motor to raise a heavy object to a certain height. In this article, we will delve into the process of determining how long it would take a 2750-Watt motor to lift a 385-kg piano to a sixth-story window located 16.0 meters above the ground.

First and foremost, let's analyze the power output of the motor in question. The motor is rated at 2750 Watts, which indicates the rate at which it can perform work. Power (P) is defined as the energy transferred or converted per unit time. In this case, the power output of the motor is 2750 Watts, i.e., 2750 Joules per second.

To raise the 385-kg piano to a height of 16.0 meters, we need to calculate the total work done against gravity. The work done (W) is given by the formula:

\[ W = m \times g \times h \]

Where:

- \( m = 385 \, kg \) (mass of the piano)

- \( g = 9.8 \, m/s^2 \) (acceleration due to gravity)

- \( h = 16.0 \, m \) (height to which the piano is lifted)

Substitute the given values into the formula to find the total work required to lift the piano.

\[ W = 385 \, kg \times 9.8 \, m/s^2 \times 16.0 \, m \]

\[ W = 60080 \, J \]

Therefore, the motor needs to provide a total of 60080 Joules of work to lift the piano to the sixth-story window.

With the power output of the motor and the total work required, we can determine the time it would take for the motor to lift the piano to the specified height. The relationship between power, work, and time is given by the formula:

\[ P = \frac{W}{t} \]

Where:

- \( P = 2750 \, W \) (power of the motor)

- \( W = 60080 \, J \) (work done)

- \( t \) is the time taken

By rearranging the formula, we can solve for time:

\[ t = \frac{W}{P} \]

\[ t = \frac{60080 \, J}{2750 \, W} \]