The Acceleration of the system is 6.41 m/s².
Given,
α= 15°, m₁ = 7kg
β= 65°, m₂ = 11 kg
Let, a be the acceleration and T is the tensions at the end it's the cord.
Let, the mass m₂ be coming down along the inclined plane along the inclined surface towards downward m₂g sin β and the tension in the upward direction,
Resultant force, m₂a=m₂g sin β -T
11a=((11) ×g sin 65°) -T ...(i)
Now, considering the motion of m₁ which moves downwards, the forces are m₁g sinα, and T both are acting downwards.
Resultant force m₁a = m₁g sin α+T
7a =7g sin 15°+T ...(ii)
Solving both the equations by adding them,
18a=11gsin 65°+7g sin 15°-T+T
18a=11gsin 65°+7g sin 15°=115.45
a=115.45/18=6.41 m/s²
Hence, the Acceleration of the system is 6.41 m/s².
Learn more about the acceleration here:
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