# 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

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?

## This Post Has 3 Comments

1. larkinc2946 says:

Explanation:

We can write Newton's 2nd law as applied to the sliding mass $m_1$ as

$T - m_1g\sin38 = m_1a\:\:\:\:\:\:\:(1)$

For the hanging mass $m_2,$ we can write NSL as

$T - m_2g = -m_2a\:\:\:\:\:\:\:(2)$

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

$(m_1 + m_2)a = m_2g - m_1g\sin38$

or

$a = \left(\dfrac{m_2 - m_1\sin38}{m_1 + m_2}\right)g$

$\:\:\:\:= 0.30\:\text{m/s}^2$

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

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

$\:\:\:\:=24.7\:\text{N}$

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3. Expert says:

terd muffin is the correct answer