Question

A 950 kg sack of cement is lifted to the top of a building 50 m high by an electric motor.

a Calculate the increase in the gravitational potential energy of the sack of cement.
b The output power of the motor is 4.0 kW. Calculate how long it took to raise the sack to the top
of the building.
c The electrical power transferred by the motor is 6.9 kW. In raising the sack to the top of the building,
how much energy is wasted in the motor as heat?

Answer 1

a. ΔU = 465500 J

b. t = 1.94 mins

c. 337487.5 J

Step-by-step explanation:

a. Use ΔU = -mgh, here m = 950 kg, g = -9.8m/s² (since velocity decreases when you go up) h = 50m

b. Given, output power, P = 4kw = 4000w

Power is rate of work done as in P = W/t

From a, we found out that 465500J work was done

So, putting the values, we get,

4000 = 465500/t

⇒t = 116.375s

⇒t = 1.94 mins [Divide by 60]

c. Here we need to work with either power or work, but if we work with power we'd later have to convert it to work.

let's proceed with work.

Given, electrical power supplied by the motor = 6.9kw = 6900w

∴ work done by the motor = 6900 × 116.375 = 802987.5 J

Work done by the motor to bring the sack up = 465500 J

The difference of the work done will indicate the lost energy

∴ΔW = 802987.5 - 465500 = 337487.5 J