Answer
(a) Please refer to the free-body diagrams.
(b) The magnitude of the upward acceleration of the bricks is $a = 2.96~m/s^2$.
(c) $T = 191~N$
The tension $T$ is 44 N greater than the weight of the bricks. The tension $T$ is 83 N less than the weight of the counterweight.
Work Step by Step
(a) Please refer to the free-body diagrams.
(b) bricks:
$T - m_1~g = m_1~a$
$T = m_1~a + m_1~g$
counterweight:
$m_2~g - T = m_2~a$
$m_2~g - m_1~a - m_1~g = m_2~a$
$a = \frac{m_2~g - m_1~g}{m_1+m_2}$
$a = \frac{(28.0~kg)(9.80~m/s^2)-(15.0~kg)(9.80~m/s^2)}{(15.0~kg)+(28.0~kg)}$
$a = 2.96~m/s^2$
The magnitude of the upward acceleration of the bricks is $a = 2.96~m/s^2$.
(c) $T = m_1~a + m_1~g$
$T = (15.0~kg)(2.96~m/s^2) + (15.0~kg)(9.80~m/s^2)$
$T = 191~N$
The weight of the bricks is $(15.0~kg)(9.80~m/s^2)$, which is 147 N.
The weight of the counterweight is $(28.0~kg)(9.80~m/s^2)$, which is 274 N.
The tension $T$ is 44 N greater than the weight of the bricks. The tension $T$ is 83 N less than the weight of the counterweight.
