Question

3. A box (mass = 100 g) is initially connected to a compressed (r = 80 cm) spring (k = 100 N/m) at point A. It was released and started moving along the horizontal surface (Plze = 0.2) until it moves up along the inclined surface (Ilk = 0.3). The box then stops at point D alone the incline. Consider that e = 30 m and 0 = 30°. (a) What is the velocity of the box at point B?) (Ans: 25.24 m/s) (b) What is the change in kinetic energy from point B to point C? (Ans.: -5.88 J) (c) What is the velocity of the box at point C? (Ans.: 22.79 m/s) (d) What is the length of the incline, f? (Ans. 34.88 m) (e) What is the change in gravitational potential energy from point A to point D? (Ans. 17.10 ) Wwwm e x + А В с D

Answer 1

(a) Let V be the velocity at AB B Then I ka = 1 mV+M mq) x { Conservation of energy nu 2 kn - 21 .24 mgx. 2 2 =) Voo ka - 2M mgx m (100) (0.8) - 210.2)(0.1) (9.8) (018) B 0.1 V B = 25. 236 m/s V = 25, 24 m/s (b) change in kinetic energy from Blec is work done against friction. AKE = (1 mg) e cos180 - eMgm =) AKE - - (30) (0.2)(9.8) 10.1) = -5.88 T AKE = -5.88J ✓

(© let Yo be the velocity at 'c' Then I m - m = -5,885, =) I mv. = I my 5.88 -) V mya (5.88)2 (0.1)(25,24)(5.88)(2) V = = 22,7916 Orl =) = 22,79 m/s incline is (1) stopping distance along the mi f= (22,7916) 2 0 + M coso) 2x9.8{Sm 36 (0.3)(CS30) 28 g (sin + =) = ✓ f = 34,881 m 34.881 m =) f- 34.88 m Ч LE) Chang in gravitational polential energy AU= figh-of-mg {fsmo] -0 = mgf smo -) AU= (0.1)(9.8) (34,88)(Sims38) - 17.09 = 17.10J ✓