A particle is moving an (X )-axis has potential energy (U=2-20 x+5 x^2 ~J ) along (X )-axis. The particle is released at (x=-3 ). The maximum value of (x ) will be ( (x ) is in metre and (U ) is in joules)

Step-by-step solution

Potential energy, (U=2-20 x+5 x^2 J ) When particle gains whole of potential energy in the form of kinetic energy at (x=-3 ~m ), then it will travel maximum distance till whole of its kinetic energy becomes zero. ( array ll i.e., & K=U=0 & 2-20 x+5 x^2=0 & 5 x^2-20 x+2=0 & x= -(-20) (-20)^2-4 5 2 2 5 & = 20 360 10 = 20 18.97 10 & x=2 1.9 array ) For maximum value of (x ), taking + ve sign ( gathered x=2+1.9=3.9 Total distance =|-3|+3.9 =6.9 ~m 7 ~m gathered )