A helicopter winches up an injured person from a boat. The person plus equipment has a mass of 104 kg. If the height of the lift is 84 m and it is carried out at a uniform rate over 70 s, what is the power expended by the helicopter's winch motor? Assume that the engine is 61% efficient.
To find the power expended by the helicopter's winch motor, we need to consider the work done in lifting the injured person and equipment.
The work done (W) is given by the formula:
W = mgh
where m is the mass (104 kg), g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height (84 m).
Plugging in the values:
W = 104 kg × 9.8 m/s^2 × 84 m
W = 86,646.4 Joules
The power (P) expended is the work done divided by the time (t), which is 70 seconds in this case.
P = W / t
P = 86,646.4 J / 70 s
P = 1,237.8 Watts
However, we need to consider the engine efficiency, which is stated as 61%. This means that only 61% of the energy input is converted into useful work.
Therefore, the actual power expended by the winch motor is:
Actual Power = P * Efficiency
Actual Power = 1,237.8 Watts * 0.61
Actual Power = 754.2 Watts
So, the power expended by the helicopter's winch motor is approximately 754.2 Watts.