There are 2 blocks attached by a cord, m1 is on a table and m2 is dangling off t

Question

There are 2 blocks attached by a cord, m1 is on a table and m2 is dangling off the side. There is a pulley wheel on the corner of the table. m1 = 9.5 kg and m2 = 4.5 kg. The coefficient of static friction between m1 and the horizontal surface is 0.60 while the coefficient of kinetic friction is 0.30.

(a) If the system is released from rest, what will its acceleration be?

(b) If the system is set in motion with m2 moving downward, what will be the acceleration of the system?


Please answer in detail

Explanation / Answer

Given m1 = 9.5 kg, m2 = 4.5 kg (a)If the system is released from rest, The net force acting on m1 is     f1-T = m1 a    µs (m1g)-T = m1 a    a = ( µs (m1g)-T)/ m1-----------(1) The net force acting on m2 is T = m2 g = 4.5 *9.8 m/s^2             = 44.1 N Now equation (1) becomes a = ((0.60 *9.5 kg *9.8 m/s^2)-44.1)/9.5 kg    = 1.23m/s^2 (b)If the system is set iinto motion , The net force acting on m1 is     T- f1 = m1 a   T- µk (m1g) = m1 a    a = ( T -µk(m1g))/ m1      = (T-27.93)/9.5-----------(1) The net force acting on m2 is m2a = m2 g-T = (4.5)(9.8)-T =44.1 -T T = 44.1 -4.5 a substitute the value of T in (1) a =( 44.1-4.5 a-27.93)/9.5 9.5 a = 16.17-4.5a 14 a = 16.17 a =3.6 m/s^2 The net force acting on m1 is     T- f1 = m1 a   T- µk (m1g) = m1 a    a = ( T -µk(m1g))/ m1      = (T-27.93)/9.5-----------(1) The net force acting on m2 is m2a = m2 g-T = (4.5)(9.8)-T =44.1 -T T = 44.1 -4.5 a substitute the value of T in (1) a =( 44.1-4.5 a-27.93)/9.5 9.5 a = 16.17-4.5a 14 a = 16.17 a =3.6 m/s^2

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