I would tell him, in the kindest, most gentle way I could manage,
to fahgeddaboudit.
The total amount of energy doesn't change. Energy is never created,
and it never disappears. If you have some energy, then it had to come
from somewhere, and if you used some energy, then it had to go
somewhere.
You can never get more energy out of the electromotor than you put into it,
and in the real world, you can't even get THAT much out, because some
of it is always used on the way through.
Pour yourself a cold glass of soda, then look up "Perpetual Motion" or
"Free Energy" on the internet, relax, and enjoy the show. They are all
fakes. They may not all be intentionally meant to fool you, but they are
all impossible.
I think the small mass Sorry if I’m incorrect
c .... work is force x distance as a definition
Answer:
a). M = 20.392 kg
b). am = 0.56 
(block), aM = 0.28 (bucket)
Explanation:
a). We got N = mg cos θ,
f = 
= 
If the block is ready to slide,
T = mg sin θ + f
T = mg sin θ + .....(i)
2T = Mg ..........(ii)
Putting (ii) in (i), we get
M = 20.392 kg
b). 
.............(iii)
Here, l = total string length
Differentiating equation (iii) double time w.r.t t, l, h and h' are constants, so
.....................(iv)
We got, N = mg cos θ
∴ 
................(v)
Mg - 2T = M
(from equation (iv))
.....................(vi)
Putting (vi) in equation (v),
![$\frac{g\left[\frac{M}{2}-m \sin \theta-\mu_K m \cos \theta\right]}{(\frac{M}{4}+m)}=a_m$](https://tex.z-dn.net/?f=%24%5Cfrac%7Bg%5Cleft%5B%5Cfrac%7BM%7D%7B2%7D-m%20%5Csin%20%5Ctheta-%5Cmu_K%20m%20%5Ccos%20%5Ctheta%5Cright%5D%7D%7B%28%5Cfrac%7BM%7D%7B4%7D%2Bm%29%7D%3Da_m%24)
![$\frac{9.8\left[\frac{20.392}{2}-10(\sin 30+0.5 \cos 30)\right]}{(\frac{20.392}{4}+10)}=a_m$](https://tex.z-dn.net/?f=%24%5Cfrac%7B9.8%5Cleft%5B%5Cfrac%7B20.392%7D%7B2%7D-10%28%5Csin%2030%2B0.5%20%5Ccos%2030%29%5Cright%5D%7D%7B%28%5Cfrac%7B20.392%7D%7B4%7D%2B10%29%7D%3Da_m%24)
Using equation (iv), we get,
I think the answer is (D)
