Question

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4. 101 (En Ph 131 - Winter 2008 Pis applied and t is in seconds. vinter 2008 - Final Exam) A 10.0 kg block is resting on a horizontal surface when a force w seconds. The variation of P with time is shown in the figure where P is in Newtons in seconds. If the coefficients of static and kinetic friction between the block and the surface are fis " and p= 0.400 respectively, and g=10.0 m/s2 calculate a) [2] The time at which the block begins to move. b) [4] The maximum velocity reached by the block. c) [4] The total time (A1) during which the block is in motion 10 ks P P, N POE-6.25r2+501 100- 0 4 7

Answer 1

(a) When the blocks starts to move it meam pe to P- elsmg -6.25 +² + 50 t = (0-6) (10) (9.8) - 6.25 + 2 + 50 t, -58-8 = o solving this quadratic equation we get to = 1.43 sec Now the block started to move and we abblied force P as -6-25ť +50 f No 2d sur 35 'N ult27 o ( for P = know first period when -6.25 +² + sot N that a = du = (-6-25 +² +507) - ses we * - elk mg m 5t² +50tl - 0.392 ole 10

-0.695 t + 25 t - 0.392479 1.43 ů – 0.625 (43 - 4433) + 25 (42_4- 4892 _ 0.392 (4-16433 Ug = 21.16 mis for Now a force works 3 seconds constant P= 35N ă = 35 sa 13.5 mis] so va a vit at Uz - 21.16+ (3-5) (3) ve 31.66 mis This is the highest speed achieved. o now the block has constant velocity , we have only deacceleration due to kinetic friction which is given by a = -lkg = -3.92 mis? so uf- uitat o= 31.66 - 3-92 tz to = 8.076 sec Total time of motion a tz +7- tie [13-65 Seel

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