Question

1) A ball of mass, m = 2.50 kg, floats on a pool of fluid, L at either end of the pool by springs, as shown in Fig.-1. The left spring has a spring constant, 1.00 m wide. It is attached to supports 1) a. 65.0N/m, and equilibrium length, L0.600m. The spring on the right has a spring constant, k2-75.0N/m, and equilibrium length, L2-0.800m. L-1.00 m e. ki LI Figure 1 -A ball of mass, m = 2.50 kg, floats on a pool of fluid, L = 1.00 m wide. It is attached to supports at either end of the pool by springs. The left spring has a spring constant, ki = 65.0 N/m, and equilibrium length, L1 = 0.000 m. The right spring has a spring constant, k2-75.0 N/m, and equilibrium length, L2-0.800m. (Not to scale.) a) The ball is initially at rest at position, z = d. It's then pulled to the center between the two supports at z =-= 0.50 m, and let go. How long after it's let go does it take to return to d? (Neglect the drag of the fluid.) b) Supposing we now consider that the fluid exerts a drag force on the ball of fo-b, where b-20.0 kg/s. If the ball is again pulled to the center atand let go, now how long does it take to return to the positio, d? c) In Part b, once the ball gets back to the position, z = d, what's the furthest it will get from d) Supposing the drag coefficient is changed to b 40.0 kg/s. If the ball is again pulled to the e) Going back to Part b with the drag coefficient given by b20.0 kg/s, supposing the z = d after that? center at aand let go, now how long does it take to return to the position, d? ball is initially at rest at z. But then it's driven with a driving force given by F- (50.0N) cos (8.00s-). After a long time has passed, how fast will the ball be traveling whenever it passes through the point where ?

Answer 1

Given that a) EOM: K_1 (x - l_1) - K_2 [(L - x) - l_2] = mx K_1 (x - l_1) - K_2 ((1 - x) - l_2) = m. dv/dt K_1 (x - l_1) - K_1 [(l - x) - l_2] integral^1_0 dt = m integral^v_0 dV {K_1 (x - l_1) - K_2 [(1 - x) - l_2]} l^2/2 = m integral^d_1/2 dx {K_1 (x - l_1) - K_2 [(1 - x) - l_2]} + L/2 = m(d - l(2) rightarrow (1) {K_1 (pi - l_1) - K_2 [(L - x) - l_2]} t = mv = m dx/dt now at equilibrium position, K_1 (d - l_1) = K_2 [(L - d) - l_2] K_1 d - K_1 l_1 = K_2 L - K_2d - K_2 l_2 K_1 l_1 - K_2 l_2 + K_2 L = (K_1 + K_2) d \r\n d = K_1 l_1 - K_2 l_2 + K_2 L/(K_1 + K_2) putting Values