A non–linear spring exerts a restoring force of F(x) = −Ax − Bx^2 if it is stretched or compressed.?
A non–linear spring exerts a restoring force of F(x) = −Ax − Bx^2
if it is stretched or compressed.
(a) Find the potential energy function for this spring.
(b) Let U(0) = 0.0 J, A = 50.0 N/m, B = 15.0 N/m2
. A block of mass 0.500 kg is attached to the spring and
pulled a distance of 1.5 m to the right on a frictionless surface. What is the speed of the block when it is
at a position of 0.75 m to the right of the equilibrium position?
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Verified answer
F(x)dx = −Axdx − Bx^2dx
PE(x) = INT(F(x)dx) = - Ax^2/2 - Bx^3/3|0 to x = - (Ax^2/2 + Bx^3/3) ANS a.
ke = PE - pe = 1/2 mV^2; so V = sqrt(2(PE - pe)/m) = sqrt(2*((50*1.5^2/2 + 15*1.5^3/3) - (50*.75^2/2 + 15*.75^3/3))/.5) = 15.1 m/s. ANS.