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A block A of mass 8 kg is placed on a frictionless horizontal table. A thread tied to it passes over a frictionless pulley and carries a body B of mass 2 kg at the other end. Find the acceleration of the system. Also find the tension in the thread. If the thread is cut into two and tied to the ends of a spring of forceconstant 1600N/m, find the amount of stretching of the spring. Neglect the mass of thread and of spring (g = 9.8 m//s^(2)) |
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Answer» Solution :Figure(a) represents the arrangement of the first PART of the problem Here `m_1 = 8kg , m_2 = 2kg ` If a is acceleration of system and T the tension in STRING  then we have for the motion of block B , `m_2 g - T = m_2 a ` ....... (1) and for the motion of blockA , T = ` m_1 a ` ....... (2) ( since weight `m_1g` is BALANCED by normal reaction R ) Adding (1) and (2) , we get ` m_2 g = (m_1 + m_2) a ` `:. ` acceleration ` a=(m_2)/(m_1 + m_2) g` ...... (3) Substituting this is (2) , we get , Tension `T=(m_1 m_2)/( m_1 + m_2 ) g `.......... (4) In second part the spring is introducedand the arrangement is shown in figure (b) . The spring is pulle by Tension `T=(m_1 m_2)/( m_1 + m_2) g ` If x is the streching of spring , then we have `T= Kx (or ) x =(T)/(K) = (m_1 m_2 g)/( (m_1 + m_2)K)` substituting ` m_1 = 8 kg , m_2 = 2kg ` . K= 1600 N/m in (3), (4)and (5) , we get acceleration ` a= ( m_2)/( m_1 +m_2)g= (2) /( 2 + 8 ) xx 9.8 = 1.96 m//s^(2)` Tension `T= (m_1 m_2)/( m_1 + m_2)g = (2 xx8)/(2 + 8) xx 9.8 = 15.68` N and stretching ` x= (T)/(K) = (15.68)/(1600) = 0.0098 m` . |
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