The input power of an electric motor is 200W. It is used to raise a mass of 10kg at constant speed.If the efficiency of the motor is 40%, through what height will the mass be raised in 1 second?A. 0.5mB. 0.8mC. 1.2mD. 2.0m

To solve this problem, we need to understand a few basic concepts:

1. Efficiency of a machine is the ratio of useful output power to the total input power, expressed as a percentage. In this case, the efficiency of the motor is 40%, which means that 40% of the input power is converted into useful work.

2. The work done by a force in lifting a mass is given by the formula: Work = mass * gravity * height. In this case, the mass is 10kg and the acceleration due to gravity is approximately 9.8m/s^2.

3. Power is the rate of doing work or the amount of work done per unit time. It is given by the formula: Power = Work / time. In this case, the time is 1 second.

Now, let's solve the problem:

Step 1: Calculate the output power of the motor. Since the efficiency is 40%, the output power is 40% of the input power. So, Output Power = 40/100 * 200W = 80W.

Step 2: Calculate the work done in 1 second. Since power is work done per unit time, the work done in 1 second is equal to the output power. So, Work = Output Power * time = 80W * 1s = 80J (Joules).

Step 3: Calculate the height. We know that Work = mass * gravity * height. So, height = Work / (mass * gravity) = 80J / (10kg * 9.8m/s^2) = 0.816m.

So, the mass will be raised through a height of approximately 0.8m in 1 second. Therefore, the correct answer is B. 0.8m.