Question

An object of mass 2.0 kg is att at its equilibrium position, it is given a sharp downward blow with a hammer so that its initibl speed is 0.30 m/s to the top of a vertical spring that is anchored to the flioor. The unstressed length of the spring is 8.6 cm and the length of the spring when the object is in is 6.1 cm. When the object is resting (a) To what maximum height above the floor does the object eventually rise? (b) How long does it take for the object to reach its maximum height for the first time? Select 1) What minimum initial speed must be given to the object for the spring to be unstressed at some time? cm/s

Answer 1

P.E_S = Rx_i/2 using F_g = F = > mg = kx k = mg/x_i Therefore P.E_s = mg x_i/2 Now, using conservation of energy E_i = E_f = > P.E_s + K.E_i = P.E_s^f + P.E mg x_i/2 + mv_i^2/2 = kx^2/2 + mg delta x mg x_i/2 + mv_i^2/2 = mgx^2/2x_i + mg (x_i - x) Solving for x: x = x_i - v_i squareroot x_i/g Therefore x_i = 8.6 cm - 6.1 cm x_i = 2.5 cm Therefore x = 2.5 times 10^-2 - 0.30 squareroot 2.5 time 10^-2/9.01 = 2.5 times 10^-2 - 1.51 times 10^-2 x = 0.99 times 10^-12 = > x = 0.99 cm Therefore h_max = 8.6 - 0.99 = > h_max = 7.61 cm t_i = 3t/4 = > t_i = 3/4 times squareroot m/R = 3 pi/ R = 3 pi/2 squareroot x_i/g = 0.238 sec No x = x_i - V_i squareroot x_i/g = 0 = > V_i = squareroot x_i g