Question

The system is initially moving with the cable taut, the 10-kg block moving down the rough incline with a speed of 0.130 m/s, and the spring stretched 25 mm. By the method of this article, (a) determine the velocity v of the block after it has traveled 108 mm, and (b) calculate the distance d traveled by the block before it comes to rest. 10 kg >K= 155 N/m Up = 0.18 7 460 Answers: (a) After it has traveled 108 mm, v = m/s (b) When it stops, d = mm

Answer 1

Given intial Speed Vo = 0.13 mis digplocement of Spring 8 = 25 mm = 0.025 m we know k = 155 MM 2F 6 /04 46 Fek6 1 = 155*00025 F = 3.875 N -0.18 → If F is the force present in the - is the force in spring then 'f' will in cable (from diagram) Now , Free body diagram of block Wema 1/46 » WCDS 46 = 10*9.8) = 98.1 N W Win46 V ma = Inertia force f = friction force = MN :. IFA FO E WOOSH46 -98.) *C0846 : 68.146 N { fy=0 =) QF+ f + ma = W sinuo 2* 3.875 + 0.1868.146 + 10*a = 98.189n46 a : 5.05 m/s2 we know constant acceleration where for linear motion with v²v. 2=cas Vfinal velocity Vo = intial velocity S: distante travelled = 108 mm = 0.108 m

=> v² 0.13² = 2x5.05x00108 =) v = 1.052 m/s b) at rest final velocity =o =) = 0 B v² Vo² = 2as 0?-0.13? = 2 x 5.05% S - s =-10673x103 m : -1,673 mm (-ve sign indicate . that block moves (0) down wora) s = 10673 mm ( clown word) ! .