Question

Problem 1 A roller coaster cart has a mass of 200.0kg, which includes its passengers. The cart is pushed off its beginning position by a spring with a spring constant of k 150.ON/m that has been compressed by Ax 0.50m from its equilibrium length. Friction is negligible everywhere, except for a 2-meter-long rough patch on the initial slide (see the figure below), where the coefficient of kinetic friction is uk 0.4. (For the purposes of calculation involving friction, here you may proceed as if the cart has negligible length compared to the length of the rough patch.) At the location of the rough patch, the slide makes an angle of 70.0 with the horizontal. The cart then reaches the bottom of a valley, 8.0 m below the starting position. At the bottom, the valley has a radius of curvature of R 3.0m. The cart then continues to climb to a peak, which is 6.0 m above the bottom of the valley. At the top, the peak has a radius of curvature of R2 2.5m. Find how many gs the passengers feel at the bottom of the valley and the top of the peak. At the top of the peak, are the passengers pressing against their seats, or against their shoulder harnesses? Δι-0.50 m 00 α 70.0%

Answer 1

Mass M = 200.0 kg, h = 8.0 m Delta x = 0.50 m K = 150.0 N/m Total energy at top starting point U_i = mgh + 1/2 k (Delta x)^2 force of friction f = mu N = mu mg cos 70 degree = 2.0 m hence work done against friction W = f d So kinetic energy at bottom of valley E_i = U_i - W = mgh + 1/2 k (Delta x)^2 - fd = 200 times 9.8 times 8.0 + 1/2 times 150 (0.50)^2 - 0.4 times 200 times 9.8 (cos 70)2 = 15680 + 18.75 - 536.29 = 15162.46 J 1/2 m v^2_1 = 15162.46 J = > v_1 = squareroot 2 times 15162.46/200 V_1 = 12.3 m/s At the bottom of valley N_1 - mg = m v^2_1/R Therefore N_1 = m(g + V^2_1/R) = 200 (9.8 + 151.62/3)N = 12068 N = 12068/g times of g = 12068/9.8 g