Answer
(a) The mass of the stone is 2.00 kg.
(b) The tension in the wire is 14.0 N.
Work Step by Step
(a) We can find the acceleration of the stone.
$y = \frac{1}{2}at^2$
$a = \frac{2y}{t^2}$
$a = \frac{(2)(12.6~m)}{(3.00~s)^2}$
$a = 2.80~m/s^2$
We can find the velocity of the stone after it falls for 3.00 seconds.
$v^2 = v_0^2+2ay = 0 + 2ay$
$v = \sqrt{2ay} = \sqrt{(2)(2.80~m/s^2)(12.6~m)}$
$v = 8.40~m/s$
We can use conservation of energy to find the mass $m$ of the stone. The magnitude of the change in potential energy of the stone will be equal to the sum of the stone's kinetic energy and the rotational kinetic energy of the pulley.
$mgh = \frac{1}{2}mv^2 +\frac{1}{2}I\omega^2$
$2mgh = mv^2 +\frac{1}{2}MR^2\omega^2$
$2mgh - mv^2 =\frac{1}{2}MR^2(\frac{v}{R})^2$
$m(2gh - v^2) =\frac{1}{2}Mv^2$
$m =\frac{Mv^2}{2(2gh - v^2)}$
$m =\frac{(10.0~kg)(8.40~m/s)^2}{2[(2)(9.80~m/s^2)(12.6~m) - (8.40~m/s)^2]}$
$m = 2.00~kg$
The mass of the stone is 2.00 kg.
(b) $\sum F = ma$
$mg-T = ma$
$T = m(g-a)$
$T = (2.00~kg)(9.80~m/s^2-2.80~m/s^2)$
$T = 14.0~N$
The tension in the wire is 14.0 N.
