Question

Two boxes (m1 = 43.0 kg and m2 = 39.5 kg) are connected by a light string that passes over a light, frictionless pulley. One box rests on a frictionless ramp that rises at 30.0^\circ above the horizontal (see the figure below), and the system is released from rest. find the acceleration of each box

Answer 1

this is what i came up with
T-mg =ma
t-mgsibthera=ma

Answer 2

Both masses have the same acceleration:
\[a=\frac{M-m\sin 30^\circ}{M+m}g\]
where m is the mass on the incline.

Answer 3

If we realize that the tension acts parallel to the surface of the incline, we can sum the forces that act parallel to the surface of the incline when we analyze \(m_1\). Let's sum the forces about \(m_1\) parallel to the incline. \[F = ma \rightarrow m_1 a = T - m_1 g \sin(\theta)\]Now, let's sum the vertical forces about \(m_2\). \[-m_2 a = T - m_2 g\]
T and a must be equal for both masses. Since we want to know a, let's solve both for T and set them equal to each other. \[T = m_1 a + m_1 g \sin(\theta)\]\[T = -m_2 a + m_2 g\]\[m_1 a + m_1 g \sin(\theta) = m_2 a + m_2 g\]We can now solve for a\[a (m_1 + m_2) = m_2g - m_1 g \sin(\theta)\]\[a = {m_2 g - m_1 g \sin(\theta) \over m_1 + m_2}\]

Answer 4

let the acceleration of the first block m1 be a hence the acceleration of second block also will be a due to both connected by same string.

hence from the figure::
m1gsin30-T=m1a...(1) T is tension
T-m2g=m2a....(ii)
from (I) and(ii)
(m1/2-m2)g=(m1+m2)a
a=(m1/2-m2)g/(m1+m2).. it will give u negative value of a .
that means ,which direction we have assumed,opposite to it acceleration direction will be..