(II) The 9.0-kg mass in Fig. is held just barely in contact...

(II) The 9.0-kg mass in Fig. is held just barely in contact with a spring for which $k = 430 \text{ N/m}$. When that mass is released, it falls, compressing the spring and pulling the 3.0-kg mass up. How far does the 9.0-kg mass fall before momentarily coming to rest? Ignore friction in the pulley.

Transcript

00:01 Here we have a mass on table which is attached to a spring and another mass is hanging and we have to find out the combined kinetic energy of the two blocks in the hanging block has fallen by 0 .09 meters so we will take we will use the work energy theorem work energy theorem where work done by all the forces on the system of the two masses is equal to change in kinetic energy of the masses so work done by gravity plus work done by spring is equal to final kinetic energy minus initial kinetic energy.

00:45 Work done by gravity is 2m which is the mass of the hanging block g h.

00:55 This is the force 2mg and h is the displacement.

01:00 So this is a work done by gravity plus spring work is minus half k x squared when the mass falls the spring is stretched by the same amount so x will be equal to h and final kinetic energy k for the system minus initial kinetic energy zero so we have 2 m g h minus half k h squared the extension of spring is same as the distance fallen by the hanging mass so kinetic energy kf is equal to 2mg h minus half k h squared equal to 2 mass is 2 kilogram g is 9 .8 h is height fallen 0 .09 meters minus half k is spring constant 200 and h square 0 .09 meter square so this is the kinetic energy equal to 2 .718 jule this is the kinetic energy of the system of the two blocks now the speed of the two blocks will be same v because they are connected by inextensible inextensible string so the total kinetic energy kf which we found will be equal to some of the two kinetic energies half m v square sorry this is half mv square for the mass on table plus half 2m v square for the mass on for the hanging mass so we have 2 .718 jule equal to mv square upon 2 plus mv square equal to 3 by 2 mv square and the value of m is 2 kilogram so upon solving we get v equal to 0 .95 chute so the kinetic energy of the hanging mass will be k2 equal to half 2m v square half 2 into 2 kilogram times point the unit of speed is meter per second not jule 0 .95 meter per second square so this gives us kinetic energy equal to 1 .812 jule for the last part we have to find out the maximum distance which the hanging block falls before momentarily stopping so so here we will again use work energy theorem, work done by gravity plus work done by spring is equal to change in kinetic energy.

04:34 But here the block stops momentarily.

04:38 So the final kinetic energy is zero and initial kinetic energy is also zero.

04:44 So changing kinetic energy is zero...

Question

(II) The 9.0-kg mass in Fig. is held just barely in contact with a spring for which $k = 430 \text{ N/m}$. When that mass is released, it falls, compressing the spring and pulling the 3.0-kg mass up. How far does the 9.0-kg mass fall before momentarily coming to rest? Ignore friction in the pulley.

Please give Ace some feedback