Question

A hoist powered by a \[15kW\] motor is used to raise a \[500kg\] bucket to a height of 80m. If the efficiency is \[80\% \]. Find the time required.

Answer 1

Hint:In order to calculate the time required, we need to use the formula of work, power and energy here. Power is the amount of energy transferred or converted per unit time. In the International System of Units, the unit of power is the watt, equal to one joule per second. In older works, power is sometimes called activity. Power is a scalar quantity. The output power of a motor is the product of the torque that the motor generates and the angular velocity of its output shaft.

Step by step solution:
Upward force needed= bucket's weight\[ = (mg) = 500 \times 9.8 = 4900N\]
Power available\[(P) = 0.80 \times 15 \times {10^3}W = 1.2 \times {10^4}W\]
Since,
\[P = W/t = Fs/t\]
So, \[t = \dfrac{{4900 \times 80}}{{1.2 \times {{10}^4}}} = 32.7sec\]

Thus, the correct answer to this question is \[32.7\sec \].

Note: The dimension of power is energy divided by time. In the International System of Units (SI), the unit of power is the watt (W), which is equal to one joule per second. Mechanical power is also described as the time derivative of work. However, it has a very specific meaning. It is a measure of the rate at which work is done (or similarly, at which energy is transferred). The ability to accurately measure power was one of the key abilities which allowed early engineers to develop the steam engines which drove the industrial revolution.