A 1.00 x 10^4 W electric motor is used to lift a 955 kg object vertically 25.0 m at a constant velocity in 45.0 s. What is the efficiency of the motor?

I did P=W/t (1.00x10^4 x 45 =W)
And got 4.50 x 10^5
Then I did
Ep' = Ek
955 x 9.81 x 25
= 234213.75
But I wasn't sure what these values meant, but I plugged them into the efficiency equation anyways
234213.75 / 450000 x100 = 52.0 % efficient.

I am just unsure of what the numbers really mean and what I am actually solving here.

In this problem, you are given the power of the electric motor (1.00 x 10^4 W), the mass of the object being lifted (955 kg), the height it is lifted vertically (25.0 m), and the time it takes to lift the object (45.0 s).

To find the work done by the motor, you correctly used the formula P = W/t, where P is power, W is work, and t is time. Rewriting the formula, we have W = P x t. You calculated the work done by multiplying the power (1.00 x 10^4 W) by the time (45.0 s) and got a result of 4.50 x 10^5 J. This represents the amount of energy transferred by the motor to lift the object.

Next, you equated the change in potential energy (Ep') to the change in kinetic energy (Ek). The change in potential energy can be calculated using the formula Ep' = mgh, where m is the mass of the object, g is the acceleration due to gravity (9.81 m/s^2), and h is the height the object is lifted. You correctly substituted the given values into the formula and obtained an answer of 234213.75 J, which represents the change in potential energy.

Now, let's discuss the efficiency of the motor. Efficiency is a measure of how effectively a device converts energy input into useful work output. In this case, it tells us how efficiently the electric motor converts electrical energy into the work required to lift the object.

Efficiency can be calculated using the formula: efficiency = (useful output energy / input energy) x 100%. Since the output energy is the work done to lift the object (4.50 x 10^5 J) and the input energy is the electrical energy supplied to the motor, we can substitute these values into the formula.

To find the input energy, we need to consider the electrical energy used by the motor. The power (P) of the motor represents the rate at which it uses electrical energy. So we can calculate the input energy by multiplying the power (1.00 x 10^4 W) by the time (45.0 s). This gives us an input energy of 4.50 x 10^5 J, which is the same as the work done by the motor.

Plugging in the values, we get: efficiency = (4.50 x 10^5 J / 4.50 x 10^5 J) x 100%. Simplifying, the units cancel out, leaving us with an efficiency of 100%. This means that the motor is 100% efficient in converting electrical energy into useful work in this particular scenario.

So, based on your calculations, your result of 52.0% efficiency seems incorrect. The correct answer is 100% efficiency since all the electrical energy supplied to the motor is converted into the work required to lift the object.

Never do calculations if you are unsure of what they mean.

eff= power out/power in

power out= 955*g*25
power in= 1E4*45

caclulate efficiency. It is 52 percent. My experiences with electric motor (smaller than this) is that they are about 30 percent efficienct.