Work Energy And Power Question 295
Question: A uniform rope of linear mass density ‘k and length (. is coiled on a smooth horizontal surface. One end is pulled up with constant velocity v. Then the average power applied by the external agent in pulling the entire rope just off the horizontal surface is
Options:
A) $ \frac{1}{2}\lambda \ell v^{2}+\frac{\lambda {{\ell }^{2}}g}{2} $
B) $ \lambda \ell gv $
C) $ \frac{1}{2}\lambda v^{3}+\frac{\lambda \ell vg}{2} $
D) $ \lambda \ell vg+\frac{1}{2}\lambda v^{3} $
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Answer:
Correct Answer: C
Solution:
[c] Initial $ KE.=0 $ , Initial $ P.E.=0 $
When the rope is just pulled off the table, final $ K.E =\frac{1}{2}(\lambda \ell )v^{2}, final P.E. = (\lambda \ell )g\ell /2 $ time taken $ =t=\ell /v $
$ Average power =\frac{net change in energy}{time} $
$ =\frac{\frac{1}{2}\lambda \ell v^{2}+\lambda \ell g\ell /2}{\ell /v}=\frac{1}{2}\lambda v^{3}+\frac{\lambda \ell vg}{2} $
BETA
