Question

Q3

QUESTION 3 The block with mass of 10 kg is attached with the spring located on the smooth plane as shown in Figure Q3. The plane has been raised to 8 =30° from horizontal surface. When the force F= 200 N is applied, the block started moving down the plane with a speed of 2 m/s. We know that when s = 0.4 m, the spring in an unstretched condition. Given the spring has a stiffness, k=300 N/m (a) What is the distances when the block stops? (15 marks) (6) Did the distants change when the angle of inclination is reduced from 8 - 30° to 8 -0°? What does the assumption need to be considered in this problem? Please show the proof of calculation (10 marks) Figure 3 The spring and manyatem

Answer 1

= - 1x300 (²-0) +200 x + 10x 9.8x1 x=0- 1 x 1082 in = 10kg - - 1 x (x² - 0) + FX + m gsino x = 0 - 2 mv Given Vo= 2m/s Espring and k=300 N/m. F=2OON 0 = 30° n distance the mass stops Applying work energy Theorein. Wspring + Wft Wgravity x mg Let after tranelling - OKE a - 150x2 + 200x + 49 x + 20 =0 Isou2- 2498 - 200 x= -0.077 1.74 Accepted va value of x=134 Aun S = x + 0.4 = (1074 + 4) to 78 m 2014m Tovo ov=o Kn fumi U x ng b Yes the distance Sneill change when angle of inclination is reduced from 30 to oo f of Applying wore enery Theorem wsping + WF = +WF = OKE - K x 2 + Fx- 0 - 1 m vg ² x300x2 – 200x -3x10x 2 = 0 Isox2 - 2000 - 2000 x=-0.09) Accepted values of x=1.43m '. The block will stop at couta=1.83 m Am 1.43m.