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A block of mass $0.80 kg $ is attached to a cord passing through a hole in a horizontal frictionless surface. The block is originally revolving at a distance of $0.30 m$ from the hole with a speed of $0.80 m/s$ The cord is pulled from below the hole, shortening the radius of circle in which the block revolves to $0.10 m$. At this new distance the speed of block is $2.40 m/s$. How much work is done by person who pulled the cord?

\[(a)\;1.50 \;J\quad (b)\;2.05\;J \quad (c)\; 4.25 \;J \quad (d)\;1.05\;J\]

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1 Answer

Tension in $1^{st}$ case $=T_1=\large\frac{mv_1^2}{R_1}$
$\qquad=\large\frac{0.8 \times (0.8)^2}{0.3}$$=1.71N$
Tension in II case$= T_2 =\large \frac{mv^2_2}{R_2}$
$\qquad=\large\frac{(0.8)(2.4)^2}{0.1}$$=46.1 \; N$
Work done = change in kinetic energy of the block in two cases
$\qquad\qquad=\large\frac{1}{2}$$m (v_2^2-v_1^2)$
$\qquad\qquad=\large\frac{1}{2} $$(0.8)[(2.4)^2-(0.8)^2]$
$\qquad\qquad=\large\frac{1}{2} $$\times (0.8) \times (5.76-0.64)$
Hence b is the correct answer. 


answered Aug 7, 2013 by meena.p
edited Feb 17, 2014 by meena.p

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