Question

Question
03
ja) Calculate the acceleration a' of the system and the tensions 1 and 12 in the strings as shown in Figure 1)
Assume that the table and the pulleys are frictionless and the string is massless and etensible)
10 kg
Table
OO SHOT ON REDMI
5 kg
AL DUAL CAMERA
Figure !​

Answer 1

Answer:

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Explanation:

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Answer 2

Answer:

Answer:-

a = 1.4 m/s^2a=1.4m/s2

T_1 = 22.4 NT1=22.4N

T_2 = 33.6 NT2=33.6N

Given :-

g = 9.8 m/s²

\begin{lgathered}m_1 = 2kg\\ M = 8kg \\ m_2 = 4kg\end{lgathered}m1=2kgM=8kgm2=4kg

To find :-

The acceleration of the whole system and the tension force.

Solution:-

Let the mass be m_1 , M, m_2m1,M,m2 of each block respectively.

LetT_1 and T_2T1andT2 be the tension force on string.

For bock m_1m1

acceleration is in upward.

→\mathsf{T_1 - m_1 g = F_{Net} }T1−m1g=FNet

→\mathsf{T_1 -m_1g = m_1a}T1−m1g=m1a

→\mathsf {T_1 = m_1a + m_1 g}-----1T1=m1a+m1g−−−−−1

For block MM

acceleration is in right side.

→\mathsf{T_2 - T_1 = F_{Net}}T2−T1=FNet

→\mathsf{T_2 -T_1 = Ma}------2T2−T1=Ma−−−−−−2

For block m_2m2

acceleration is in downward direction.

→\mathsf{m_2g -T_2 = m_2a}m2g−T2=m2a

→\mathsf{-T_2 = m_2a - m_2 g}-----3−T2=m2a−m2g−−−−−3

Adding equation 1 , 2 and 3.

We get,

→\mathsf{ T_1 + T_2 -T_1 -T_2 = m_1a +m_1g + Ma + m_2a -m_2g }T1+T2−T1−T2=m1a+m1g+Ma+m2a−m2g

→\mathsf{0 = m_1g -m_2g +m_1a + m_2a + Ma}0=m1g−m2g+m1a+m2a+Ma

→\mathsf{m_2g -m_1g = a(m_1 + m_2 + M) }m2g−m1g=a(m1+m2+M)

→\mathsf{g(m_2 -m_1) = a(m_1 +m_2 + M)}g(m2−m1)=a(m1+m2+M)

→\mathsf{a = \left(\dfrac{g\left(m_2-m_1\right)}{m_1 + m_2 +m_3 }\right)}a=(m1+m2+m3g(m2−m1))