Question

F (N) 10 marks 10 kg 50N--- 4 8 12 The 10-kg box, which was initially at rest, is subjected to the horizontal force F, where the graph shows the variations of the magnitude of F throughout the entire period of its application from the beginning to the end for a total of 12 seconds. The coefficient of static friction and the coefficient of dynamic friction between the box and the surface are the same and their common value is 0.2. Determine how fast the box is moving at the end of the period of application of the horizontal force, i.e. at t = 12 seconds. Take g = 9.81 m/s2 (Correct to 1 decimal place.)

Answer 1

m= loky g= 9.81 m Ms = MK = 0.2 SON - - - - 0 Initially block is at rest (v=o). on 11 force f(t) is applied on block the will not move untill the friction reaches its maximum value. maximum friction is (Fs) max = MsN =O (Fs) mor = Ms my time = 0.2 (10kg) (9.81 mm) 9N Cformare = (Esl mase = 19.62N equation of force in time interval o to 4 is F(t) = kt k k is constant at t = 4 F (t)=50 50=k(4) na k = 12 .5 = F(t) = 12.5t oct 24

Calculating time when F(t) = Fs) max D 19.62 = 12.5t t= 1.5696 sec. Hence at t = 1.5696 block starts moving. at t= 1.5696 v=0. considering time interval 1.5696£ t£ 4 Ft) + 9 -DJ mg where fic is kinetic friction fl L von N * N = mg , fx = M N = Mang = (0.2) (1) (9.80) { Fnet = ma f ic = 19.62 -D F(t)- f = ma = la.st - 19.62=ma -D 9 = 12.54-19.62 = 12.St-19.02 m Ta= 1.25t-1.962 1.5696 Lt 4 du at = 1.25t-1.902 t= 4 2st - 1.962 t = 1.5696 ) dt. o (4-0) = [1.25 +2 _ 1.962 t] to JI.5696

V4 = 1.25 (42-45697) - 1.96264-1.5096) 16) - 1.96204-1.5696) V = 8. 4602 – 4.7684 V4 = 3.6917 m hence at t= 4 velocity VA is= 3.6917 m. 11 considering time interval 4 Lt L 8. F(t) = 50 fic = ung = 19.62 N Flti-fc = ma = 50 - 19.62 = 10(a) =D a = 3.038 m S2 = Ne dv = 3.038 at t= 8 du= 33.038 dt v t=4 Ve. V = 3.038 (8-4) No - 3.6917 = 12.152 V = 15. 8437 t=8 velocity is at 15.8437 m

Considering time interval 8 th 12 let force ear is F(t) = At + B where A and B are constant at t=8 F(t) = 50 50 = 8 A+B 0 at t= 12, F(t) = 0 0= 12A + B = B = -12 A 40 - By 0 and 50 = 8A-12A = A= - 12.5 B = - 120-12.5) = B = 150 = F(t) = -12.5t +150 a = FLt) F(t) - fic m -12.5t + 150-19.62 = 10 a = -1.25tt 13.038

a = -1.25 t + 13.038 du= – 1.25t + 13.036 dt t=12 t=8 25t + 13.036) at ve te * Vier Vo = -1,25 [12²-82] + 13.038 (12-8) -50t 52.154 = Vie -15.8437 = 2.184. = Vie = 17.9977 VIR ~ 18 m at t= 12 sec velocity hence is 18 m ss