Efficiency of an Electric motor pulling a truck up an inclined plane

[Aaryan34532]

Homework Statement

See picture.

Homework Equations

The Attempt at a Solution

I tried at first just doing (1/3)*(48kJ) to get the energy dissipated, but that would i=give me 16kJ

 

[jbriggs444]

How much electrical energy did the electric motor use, all told?

 

[Aaryan34532]

@jbriggs444 I figured it out.. so seeing 2/3 that's just an efficiency.. knowing that 48 is the total work out that means 48 is 2 and 1 is 24 thus 3 is 72 so then 2/3 is also 48/72 and then the energy dissapated from 72kJ to 48kJ is 24kJ so there we go. my explanation sucks, but i know it now..

 

[CWatters]

Are you saying the answer is 24kJ? That's not correct.

Write down the equation for the efficiency of a motor in terms or power out and power in.

 

[Aaryan34532]

@CWatters But the answer for this question as shown in the picture is "B" which is 24kJ and my working proves it too... what makes you say 24kJ is not correct?

 

[CWatters]

So you are saying that the motor consumes 24kj but produces 48kj. Where can I buy one :-)

 

[CWatters]

Forget that my mistake. It asked for the dissipation. So you are correct.

 

[Devank]

Homework Statement

See picture.View attachment 234970

Homework Equations

The Attempt at a Solution

I tried at first just doing (1/3)*(48kJ) to get the energy dissipated, but that would i=give me 16kJ

I have a simpler explanation. Energy dissipated = energy wasted. So, using the formula of efficiency, the total energy used by the electric motor is 72kJ. Since the truck only gained 48kJ, wasted energy = 72-48 = 24kJ(B).