Question

A spring on a horizontal surface can be stretched and held 0.6 m from its equilibrium position with a force of 66 N. a. How much work is done in stretching the spring 5.5 m from its equilibrium position? b. How much work is done in compressing the spring 4.5 m from its equilibrium position? a. Set up the integral that gives the work done in stretching the spring 5.5 m from its equilibrium position. Use increasing limits of integration (Type exact answers) The amount of work done is (Simplify your answer)

Answer 1

Since the required force of 669 to stretched o.im giren by FluJ=ka 66 = 0.6k ② K = 110 mm from The work required to stretch the spring do to x = 5.5m given by (5.5 F(x) dx Wow= 1505 110xda 110 x2/5.5 = 1663.75J work required to compress the x 50 to x = 4.5 given by spring from W (2) = -4.5 64. Flasda - 4.5 11 $ponda 21-4.5 at 2 1113.755