Q.

The 50 kg block at A is mounted on rollers so that it moves along the fixed horizontal rail with negligible friction under the action of the constant 300 N force in the cable. The block is released from rest at A, with the spring to which it is attached extended an initial amount x1= 0.233m.  The spring constant is K= 80N/m. Calculate the velocity (in m/s) of the block as it reaches position B.

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Detailed Solution

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U12=0.2331.43380x  dx=40x2|0.2331.433=80.0J

The work done by string is 300(0.6) = 180J. We now apply the work-energy equation to be system and get 80.0+180= 12 (50) (υ20) , υ= 2.0 m/s

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