Answer
(a) The tension in the rope is 38.4 N.
(b) The mass of block B is 26.7 kg.
Work Step by Step
(a) We can find the acceleration of block A.
$x = \frac{1}{2}at^2$
$a = \frac{2x}{t^2} = \frac{(2)(18.0~m)}{(5.00~s)^2}$
$a = 1.44~m/s^2$
We can use a force equation to find the tension $T$ in the rope.
$\sum F_x = ma$
$F-T = ma$
$T = F-ma$
$T = (60.0~N)-(15.0~kg)(1.44~m/s^2)$
$T = 38.4~N$
The tension in the rope is 38.4 N.
(b) Since the blocks are connected, the acceleration of block B is also $a = 1.44~m/s^2$. We can use a force equation to find the mass of block B.
$\sum F_x = ma$
$T = ma$
$m = \frac{T}{a} = \frac{38.4~N}{1.44~m/s^2}$
$m = 26.7~kg$
The mass of block B is 26.7 kg.
