Question

Step 1 The free-body diagrams of the two objects in this system are shown below. Note that the accelerations of the two objects have the same magnitude, a, with the acceleration of the object of mass m1 directed horizontally to the right and the acceleration of the object of mass m2 directed vertically downward arta ay-a 1n Since m2 is observed to drop downward 1.00 m in 1.43 s when released, the magnitude of the acceleration is found by solving the following equation for ay 2 From the diagram, we know ay-a. Substituting the given value and paying close attention to the signs, we have 1.43 s 2 and solving for a, a = 0.978 79 v > 0.979 m/s2.

Answer 1

$\Delta$ y = -1 m and v₀ = 0

from    $\Delta$ y = v₀t + 0.5a_yt²

-1 = 0 - 0.5 a (1.43)²

aceelaration a = 1/0.5x 2.045 = 0.978 m/s²

from second free body diagram

$\Sigma$ F_y = m₂a_y = m_2g - T

i.e T = m₂g - m₂a_y

=  m₂g + m₂a

= m₂ ( g + a )

if m₂ = 7 kg

T₂ = 7 ( 9.8 + 0.978 )

= 75.45 N