[albear]

A winch in a boatyard uses a 230V electric motor to raise objects from a boat onto the quay side, the winch takes 22 s to raise an object of mass 160kg through a height of 5.0m.
the motor current during this time is 14 A . Calculate the efficiency of the winch.

I know that efficiency=eng used/eng provided

eng used would be Pe=mgh =9.8*160*5=7.85*10^3 kg

I think energy provided would be electrical energy ( because that's the only other energy energy involved right)

now I know that E=0.5QV

which I thought would equal 0.5*14*22*230 (14*22 because Q=t*I)

the mark scheme however dissagrees with me removing the 0.5 from E=0.5QV

[albear]

Hello

[albear]

Hello

[albear]

I would like an answer asap if anybody can help me

[albear]

Anybody there

[ebaines]

Your "energy used" calculation is correct, excet the units isn't Kg (that's mass, not work). I would have called it "work done" and the units is Kg(meter^2)/(sec^2), or Joules.

The energy that is put into an electrical device (measured in Watt-sec) is Volts x Amps x time. So it's 230V x 14A x 22s = 70,840 watt-sec. It turns out that 1 watt-sec = 1 Joule, so your units are now consistent and all you need to do is divide the work done by this amount.