A block of mass 5 ~kg moves along the x-direction subject to the force F=(-20 x+10) N , with the value of x in metre. At time t=0 ~s , it is at rest at position x=1 ~m . The position and momentum of the block at t=( / 4) s are

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Correct answer: C. 0.5 ~m ,-5 ~kg ~m / s

F =-20 ( x - 1 2 )=-20 X ( X = x - 1 2 ) Particle will perform SHM about x = 1 2 with =2 rad / sec T = sec . Phase covered in t= 4 second =90^ . Given particle is at rest at x =1 ~m x =1 is extreme position. In 4 sec, it will be at equilibrium x =0.5 ~m and momentum = m A =5 2 0.5=5 ~kg ~m / s Direction will be towards -ve x . Hence option (3)

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