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I really need help with this problem, me and my study group were working on this

ID: 1691302 • Letter: I

Question

I really need help with this problem, me and my study group were working on this problem today but we were having problems solving it. I need to find the acceleration of the 17kg block and the tension of the connecting string. If you could show how you arrived at your answer it would be extremely helpful, thank you if you help! I really need help with this problem, me and my study group were working on this problem today but we were having problems solving it. I need to find the acceleration of the 17kg block and the tension of the connecting string. If you could show how you arrived at your answer it would be extremely helpful, thank you if you help! A 17 kg block with a pulley attached slides along a frictionless surface. It is connected by a massless string to a 4.9 kg block via the arrangement shown. The acceleration of gravity is 9.81 m/s^2. Find the acceleration of the 17 kg block. Answer in units of m/s^2. Find the tension in the connecting string. Answer in units of N. I really need help with this problem, me and my study group were working on this problem today but we were having problems solving it. I need to find the acceleration of the 17kg block and the tension of the connecting string. If you could show how you arrived at your answer it would be extremely helpful, thank you if you help!

Explanation / Answer

      according to figure the string is inextensiable and pully is frictionless so the tention T at both end is same . here two forces are acting on mass m2 one is weight w =m2g   accting verticaly downward and tention T in string accting uoward since m2g > T so                  downward force - upwrd tention = net force                                     m2g -   T      =   F2     F2 is the force which produces acceleration a in mass m2 . so                   F2 = m2a                                    m2g   - T = m2a      ............... 1     The forces accting on mass m1 are       a) its weight m1g acting vertically downward        b) reaction Z of the horizontal surface on block actind vertically upward.        c) tention T in the string .     as there is no motion perpendicular to the horizontal so Z = m1g     hence the only net force moving the mass m1 is T , the tention of the string;                         net force F1 = T      now                F1 = m1a          so                   T = m1a         .....................2     Now adding equation 1 & 2 , we get          m2g - T + T = m2 a + m1 a           m2g = a ( m1 + m2 )               a   = m2g / m1 +m2 putting values                 a =   4.9kg x 9.8m/s2 / 4.9kg + 17 kg                 a = 2.19 m/s2     Calculation of tention T .       since T = m1 a     putting a = m2g / m1 + m2        we get;            T = m1 m2g / m1 +m2     so          T = 17kg x 4.9kg x 9.8m/s2 /   17kg +4.9kg                   T = 37.3 N      hope it will help     Calculation of tention T .       since T = m1 a     putting a = m2g / m1 + m2        we get;            T = m1 m2g / m1 +m2     so          T = 17kg x 4.9kg x 9.8m/s2 /   17kg +4.9kg                   T = 37.3 N      hope it will help