Question

College of Engine X Activity ReadAndi x o lama x delection cox + - Portal x Dermer X Microsoft Power X S olutions mides X → C O File C./Users/hp/Download/Midsemester%20test%202019(V).html MOTOR hangs vertically with one end fixed and a 200 kg block attached to its other end? 7 Marks) Question 3. (15 marks) The 2 kg piece of putty is dropped 2 m onto the 18 kg block initially at rest on the two springs, each with stiftness k which adheres to the block upon contact. 1.2 kN/m. Calculate the additional defection of the springs due to the impact of the putty 18 kg - 1 UN OAShot on OnePlus Powered by Dus Carrera - Kraige and J.N. Rolton, Dynamics, 2016

Answer 1

CL sol Given :- weight of the Putty, Wp = 2 kg weight of the Block Wo = 18 kg spring stiffness for each spring , k = 1.2 kN/m. Initial height of the Putty, h = 2 m 2kg The Putty (Wp = 2 kg) strikes the block with a velocity ; vp = Jag h = /2C9-81) (2) Vp = 6.26 4 m/sec. 18 kg WW consecuestion of limeat momentum gives ; mp vp = (mpt mov' K=102 KN/m y = . v = mp VP = Wp Vp motmo Wp + Wb 2 (6.264) 2 + 18 0.6264 m/sec. V = Since K = 1.2 kN/m & there are two springs, the initial- speing deflection is ; So = Wo - 181 2K 2(1200).. S0 = 7.5 X 100 m

Data: ΡΑ ΠΟ.. Constration of energy gavess ATtA Vg to Ve = 0 0 - 1 (mot mi)/2 +0 - (mp + m), 5 + 1 mp k [CSo+ 5)2 = 502) 7=0 2 go81 -1 2+18 (0-6264)? - (2 718) S. (1200) [07.5 X60-3+5)2 - (7.5X108) -0.4 - 208 + 12001 $2 + 0.015 S T = 0 12008²-28 -0.4 =0 82-(1:67X10-3) 5 - (3.33 X 10^4) = 0 - ☺ Si-2 = -6 £ J 6 2-4 ac. za From equation (1) Equate to this formula , and you can find the value of additional difflection s of the springs due to impact of Putty.