Motor efficiency lifting 5000kg load 13m in 20s

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a motor furnishes 120 hp (746W=1hp) to a hoisting device that lifts 5000kg load to height of 13.0 meters in a time of 20.0 seconds. Find the efficiency...please help sum1 at least a formula?
 

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Energy = power * time. Figure out the ratio of the energy converted in lifting the 5000kg mass to the total energy that was used by the machine.
 
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To calculate the efficiency of a motor lifting a load, we can use the formula:
Efficiency = (Output power / Input power) x 100%
In this case, the output power is the work done by the motor, which can be calculated as:
Work = Force x Distance
Since the load is lifted to a height of 13.0 meters, the work done by the motor is:
Work = 5000kg x 9.8m/s^2 x 13.0m = 637,000 Joules
The input power is the power supplied by the motor, which is given as 120 hp. To convert this to watts, we can use the conversion 1 hp = 746 watts.
Input power = 120 hp x 746 watts/hp = 89,520 watts
Thus, the efficiency of the motor can be calculated as:
Efficiency = (637,000 Joules / 89,520 watts) x 100% = 71.1%
Therefore, the efficiency of the motor lifting a 5000kg load to a height of 13.0 meters in 20.0 seconds is approximately 71.1%.