Question

490 The125 lb block, initially at rest is subjected to the horizontal force P, which varies as shown in the graph. The coefficients of friction between the block and the horizontal surface are up = 0.4 and js = 0.5. Find a) when the block will start to move; b) when the block will stop; and c) the maximum speed that the block obtains and when that occurs. The equation of the force curve from 0 to 12 seconds is P = 3t2 - t. 420 350 (sqi) 280 210 Force 140 70 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Time (s)

Answer 1

i AN e MN W a) when -block Starts to move P z N. (NEW) : P = UW 20.5 (125) 3 62.5 34²-t = 62.5 3t'-t -62.5 =0 :t=4.734 see Ans. a) after that blocks will move in Horizoncal x direction EFx = ma P = l, N=m .86 PCT) -0.4(125) = 125 a 32,2 :.JPCA) 50 = 3.882Q a=0 * Now for Cblock doesn't ma t-o to t= 4.734 see Vzo d = = 4.734 to t= 12 sec PCH) =(3+2-4) t. 12 tot 15 P(A) = 200420 PCA) = 1680s sat t = 15 to t=20 Crat t=15 p = 2,20 ta20 P=O) max stored when t 15 to 20 1650 -suit-so 32 X= 19.405 see sans 0 Та

in (t=4734 734 to t=12 t=12] Region Not dv It a= va fact V- PCA) - so 3.582 t .. (3x² d) dt - S 50/4.1 3.852 4.734 4.734 t 2 + iť it ho soty, 1 3.85 4.734 Ja 46.734 It?!? - 94.88441 - 504-236. 77 co V(t) = 3.852 when t: 9,473 sec for max. 12 v(t) velocity fizit - it) it frodit - ito da) (( 420-50)** 4 4.734 3.582 7 3.3582 azt? 19.405 now s Cleso (reso-sit-so) dt 3.8821 12 156.6087285.935+3109.26*3-2516.04 1065.7625 ft/s v max