A body with mass  2kg moves in one direction in the presence of a force which is described by the potential energy graph passing through (2,6) as shown. If the body is released from rest at  x=2m, then it speed when it crosses x=5m is:

From x=2m to x=3.5m From, y=mx+cslope=0-63.5-2=-4y2-00-3.5=-4⇒y2=14U1=−4x+14F1=−dU1dx=+4NW1=+4×(3.5−2)=+6J From, x=3.5m to x=4.5mslope=2-04.5-3.5=2y2-00-3.5=2⇒y2=-7U2=+2x-7F2=−dU2dx=−2NW2=−2×(4.5−3.5)W2=−2JWork done from  4.5 m  to  5 m=012mv2=W1+W212×2×v2=6−2=4⇒v=2m/s