question phys

[VandyDerm]

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An elevator motor with a 3000 W output provides a constant force that transports the carriage up in 30 seconds.

Employees at the mine had several suggestions for making the operation more efficient by replacing the original elevator and surface ramp system.

Suggestion #1:
Make use of an inclined plane rather than an elevator to pull the carriage up to the surface. In this scenario, the elevator’s vertical shaft is replaced by an inclined plane 30° above the horizontal that connects the mine and the surface. A different motor supplies the force necessary to move the carriage to the surface through the incline at a constant velocity.

Suggestion #2:
Carriages are attached to a single cable that winds from the loading point to an unloading point on the surface. An empty carriage going down provides part of the force necessary to raise a full carriage coming up, while a motor supplies the remainder of the force.

Note: The mass of an empty carriage is 250 kg. The maximum gold each carriage can hold is 100 kg. The coefficient of friction between the inclined plane system and the carriage is 0.2.
Question #0001

How much energy is lost to friction if a 3000 W motor pulls the carriage up the incline?
A 0.24kJ
B 2.4kJ
C 24kJ
D 240kJ

 

[mdhopefully23]

What's the right answer?

 

[VandyDerm]

the answer is C, please explain, isnt this proper reasoning:
I forgot to mention that height is 20m

uK*mgcos30*dsin30 = total work done by friction or thermal energy. So dsin30 = 40m
Isn't the answer: 0.2*(350kg)(9.8)(cos30)(40m) = which greatly exceeds any answer up there? Where did I go wrong?

 

[dshnay]

I'm not sure if this is correct but it does bring me to your answer C

We know Power = Energy/time
so Energy = Power * Time = 3000 * 30sec = 90kJ

to lift the the elevator or the carriege (I fergot what it was) 20 meters we use W=mgh so 350*10*20 = 70kj

Subtract the difference which is attributed to friction 90-70 = 20kj

Also I think when you're trying to find friction on the incline plane you're not supposed to use dsin0, all it is Fric=Uk N = Uk mgcos0.

 

[VandyDerm]

Well, I didnt mean dsin0 i meant sin(30) = 20/Length carriage moves across plane so lenght = 40m

 

[travelbug73]

the answer is C, please explain, isnt this proper reasoning:
I forgot to mention that height is 20m

uK*mgcos30*dsin30 = total work done by friction or thermal energy. So dsin30 = 40m
Isn't the answer: 0.2*(350kg)(9.8)(cos30)(40m) = which greatly exceeds any answer up there? Where did I go wrong?

You got it right, I'm not sure why you think you are wrong. The final value if you finish your calculation is approimately 24000J = 24kJ = Answer C. Maybe you missed the kJ part since your answer is in joules and you think you got an answer that is 1000x what you should have?