A block of mass 10.2 Kg is moving horizontally with a constant velocity of 4.428 m / s . It encounters an unstretched spring with a spring constant of 200 N /m and also the floor stretch BC ( as shown ) having a coefficient of friction 0.5, simultaneously. The length of the stretch BC is 1 m . Find the distance of the point where the velocity of the block becomes zero for the first time, measured from C.

a ) 0.78 m
b ) 0.53 m
c ) 0.22 m
d ) 0.15 m

Let x be the elongation of the spring ; M , the mass of the block ; N_a the normal reaction on the block ; k the coefficient of spring and m the coefficient of friction.

Since there is no vertical motion, the weight of the block must equal its normal reaction.

   N_a = M g = 10.2 ( 9.81 ) = 100.06 N .
and the frictional force is given by ( m N_a ) = 50.00 N

From the free body diagram of the block, its equation of motion is given by

   - m N_a - k x = M a
substituting the values and rearranging to get the acceleration, a , as

   10.2 ( a ) = - 50 - 200 x

Solving the quadratic equation in x we get x = 0.78 m
Hence  distance  measured from  point C is = 1 - x  = 0.22 m.