19 point please please answer right need helpBlock on an inclineA block of mass m1 = 3.9 kg on a smooth inclined plane of angle 38is

19 point please please answer right need help Block on an incline
A block of mass m1 = 3.9 kg on a smooth inclined plane of angle 38is connected by a cord over a small frictionless
pulley to a second block of mass m2 = 2.6 kg hanging vertically. Take the positive direction up the incline and use 9.81
m/s2 for g.
What is the tension in the cord to the nearest whole number?

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  1. Explanation:

    We can write Newton's 2nd law as applied to the sliding mass [tex]m_1[/tex] as

    [tex]T - m_1g\sin38 = m_1a\:\:\:\:\:\:\:(1)[/tex]

    For the hanging mass [tex]m_2,[/tex] we can write NSL as

    [tex]T - m_2g = -m_2a\:\:\:\:\:\:\:(2)[/tex]

    We need to solve for a first before we can solve the tension T. So combining Eqns(1) & (2), we get

    [tex](m_1 + m_2)a = m_2g - m_1g\sin38[/tex]

    or

    [tex]a = \left(\dfrac{m_2 - m_1\sin38}{m_1 + m_2}\right)g[/tex]

    [tex]\:\:\:\:= 0.30\:\text{m/s}^2[/tex]

    Using this value for the acceleration on Eqn(2), we find that the tension T is

    [tex]T = m_2(g - a) = (2.6\:\text{kg})(9.51\:\text{m/s}^2)[/tex]

    [tex]\:\:\:\:=24.7\:\text{N}[/tex]

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