The magnitude of the force required to compress an imperfect horizontal spring a distance x is given by f=kx+bx3 where k = 120 n/m and b = 18 n/m3. if you use a 3.5-kg block to compress the spring a distance of 2.2 m and then let it go, what speed will the block have when it leaves the spring? assume the surfaces are smooth enough that any frictional forces can be neglected.
k = 120 N/m, b = 18 N/m³, m = 3.5 kg, xi = 2.2 m, xf = 0 m. [tex]f \: = \: kx \: + \: b {x}^{3}[/tex] Integrating both the sides by dx, we get, W = [tex]120( \frac{ {x}^{2} }{2} ) \: + \: 18( \frac{ {x}^{4} }{4} ) [/tex] x from 2.2 to 0 m, W = -395.8152 J = ∆K.E. = -0.5mv² v = √(2×395.8152/3.5) v ≈ 15.03929 m/s.
