Question

block A with a mass of 10 kg rests on a 30 degree incline. the coefficient of kinetic friction is 0.20. theattatched string is parallel to the incline and passes over amassless frictionless pulley at the top. block B with a massof 8.0kg is attached to the dangling end of the string. theacceleration of B is:
a. 0.69 up
b. 0.69 down
c. 2.6 up
d. 2.6 down
e. 0

Answer 1

Answer:

Please find attached

Explanation:

Answer 2

The acceleration of the block B is 0.69 m/s² downwards in the direction of block B.

The normal force on each block is calculated as follows;

$F_n_ A = mgcos \theta\\\\F_n_ B = m_ B g$

The frictional force on block A  is calculated as;

$F_f = \mu_k F_n\\\\F_f = \mu_ kg mgcos \theta$

The horizontal force on block A is given as;

$F_x = mgsin\theta$

The tension on the string due to each block is given as;

$T_ A = m_ A a\\\\T_ B = m_ B a$

The net force on the block B is calculated as;

$m_Bg - (T_A + m_Agsin\theta + \mu mgcos\theta) = T_B\\\\m_Bg - m_Agsin\theta - \mu mgcos\theta= T_B + T_ A\\\\m_Bg - m_Agsin\theta - \mu mgcos\theta = a(m_ B+ m_ A)\\\\a = \frac{m_Bg - m_Agsin\theta - \mu mgcos\theta}{m_B + m_ A} \\\\a = \frac{(8)(9.8)\ -\ (10)(9.8)(sin30)\ -\ (0.2)(10)(9.8)(cos30)}{8 + 10} \\\\a = 0.69 \ m/s^2 \ (in -direction \ of \ block \ B)$

Thus, the acceleration of the block B is 0.69 m/s² downwards in the direction of block B.

Learn more here:https://brainly.com/question/15412658