Question

Two blocks of 6 kilogram and 4 kilogram are placed on a horizontal frictionless surface. These two blocks are connected together with spring of spring constant 100 newton/meter. At some instant block of 6 kilogram is having speed of 10 meter/second towards positive x direction and block of 4 kilogram is moving with 10 meter/second towards 6 kilogram block. What will the kinetic energy of the system.

Answer 1

SINCE CHANGE IN K.E + CHANGE IN P.E = WORK DONE BY NONCONSERVATIVE FORCE+WORK DONE BY EXTERNAL FORCE

SINCE THERE IS NO NON CONSERVATIVE FORCE OR EXTERNAL FORCE THEREFORE CHANGE IN K.E + CHANGE IN P.E=0

 SINCE THE MAXIMUM COMPRESSION WILL OCCUR WHEN BOTH THE BLOCK STOPS OR FINAL K.E =0

 CONSIDER EXTENSION ON LEFT SIDE IS ''X'' AND RIGHT SIDE IS ''X*'' THEREFORE TOTAL EXTENSION IS X+X*

 NOW 3X(-)+6X*(+)=0 THEREFORE X=2X*..........................(1)

SINCE POTENTIAL ENERGY OF A SPRING IS (1/2)Kx2

  THEREFORE FROM CHANGE IN K.E + CHANGE IN P.E =0

 WE GET FINALK.E - INITIALK.E + FINAL P.E -INITIALP.E=0

           (0)-[(1/2 * 3 * 12)+(1/2 * 6 * 22)]+[(-1/2 * K *X2)+(1/2 *K * (X*)2)]-(0)]=0............................(2)    NOTE: ALSO MENTION THE SIGN OF THE DIRECTION 

 

     SOLVE EQUATION (1) AND (2) BY SUBSTITUTING (1) IN (2) AND GET YOUR ANSWER WHICH WILL BE X+X* 

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