Tarzan swings on a 30.0- m-long vine initially inclined at an angle of 37.0 degr

Question

Tarzan swings on a 30.0- m-long vine initially inclined at an angle of 37.0 degree with the vertical. What is his speed at the bottom of the swing if he starts from rest? If he pushes off with a speed of 4.00 m/s? Two blocks are connected by a light string that passes over a frictionless pulley as in Figure P5.38. The system is released from rest while m2 is on the floor and m1 is a distance h above the floor. Assuming m1 > m2, find an expression for the speed of m1, just as it reaches the floor. Taking m1 = 6.5 kg, m2 = 4.2 kg, and h = 3.2m, evaluate your answer to part (a), and (c) find the speed of each block when m1 has fallen a distance of 1.6m. Figure P5.38

Explanation / Answer

m1g- T =m1a T-m2g = m2a b) adding above 2 eqns we get : m1g-m2g = (m1+m2)a a = g(m1-m2)/(m1+m2) v= at = g(m1-m2)t/(m1+m2) v^2 = 2ah v = sqrt ( 2gh(m1-m2)/(m1+m2) m/s b) v =3.67 m/s c) v1 =v2 = 2.5963 m/s

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