BMAT 2014 Q 27 power output

hi
the answer is D)
i did v=10/25=0,4m/s
P=Fv=20 * 10 * 0,4 = 80W
What did i do wrong?

Hey, this is how i did it;
when the mass is being elevated it gains gravitational potential energy.
P.E=mgh
20 * 10 * 10=2000J
Power=work/time ,
work is the energy transferred
when h=10m ,t=20
so, P= 2000/20 =100W
I dont know if this is the correct way. If it isn’t, please let me know .

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Isn’t the time is 25 when the height is 10m?
Juliette, there is nothing wrong with your calculation i guess

hey, I’m really sorry i put the wrong answer here. I checked it again and this is what i found, look at the constant region of the graph. since a change in height of 10m (5-15) takes a time of 20 seconds (15-35), that’s the reason i took the time as 20 sec for a height of 10m.

Ohh i see u took the the constant slope instead of the literal 10m. I think that’s also a good idea.
I guess we can also make approximation about the final answer like Juliette got 80 W which is quite close to 100W in the given, right?

Hey, I think so that you can use the same method. Please confirm it anyway.