SOLUTION T 1 + U 1-2 = T C(31.5)(2.5)A10 6 B(0.2)D = 1 2 (7)(v 2) 2. v 2 = 2121 m>s = 2.12 km>s. Ans. (approx.

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1 4 5. When a 7-kg projectile is fired from a cannon barrel that has a length of 2 m, the explosive force exerted on the projectile, while it is in the barrel, varies in the manner shown. Determine the approximate muzzle velocity of the projectile at the instant it leaves the barrel. Neglect the effects of friction inside the barrel and assume the barrel is horizontal. F (MN) The work done is measured as the area under the force displacement curve. This area is approximately 3.5 squares. Since each square has an area of 2.5A0 6 (0.2), s (m) T + U -2 = T C(3.5)(2.5)A0 6 (0.2)D = 2 (7)(v 2) 2 v 2 = 22 m>s = 2.2 km>s (approx.)

2 4 2. Design considerations for the bumper on the 5-Mg train car require use of a nonlinear spring having the loaddeflection characteristics shown in the graph. Select the proper value of k so that the maximum deflection of the spring is limited to 0.2 m when the car, traveling at 4m>s, strikes the rigid stop. Neglect the mass of the car wheels. F (N) F ks 2 s (m) (5000)(4)2 ks 2 ds = 0 L k (0.2)3 3 = 0 k = 5.0 MN>m 2

3 4 20. The steel ingot has a mass of 800 kg. It travels along the conveyor at a speed v = 0.5 m>s when it collides with the nested spring assembly. Determine the maximum deflection in each spring needed to stop the motion of the ingot. Take k A = 5kN>m, k = 3kN>m. 0.5 m 0.45 m k k A A C Assume both springs compress T + U - 2 = T 2 2 (800)(0.5)2-2 (5000)s2-2 (3000)(s )2 = s 2-500(s 2-0. s ) = 0 s s = 0 s = m m s A = m s = m (O.K!)

4 4 38. The skier starts from rest at A and travels down the ramp. If friction and air resistance can be neglected, determine his speed v when he reaches. Also, find the distance d to where he strikes the ground at C, if he makes the jump traveling horizontally at. Neglect the skier's size. He has a mass M. Given: M h h 2 75 kg 50 m 4m 30 deg Solution: Guesses v m t s d m s Given Mg h h 2 v t d 2 Mv 2 v t dcos h.2 d sin Find v t d t s v 30.0 m d 30.2 m s = 2 gt2

5 4 53. The material hoist and the load have a total mass of 800 kg and the counterweight C has a mass of 50 kg. At a given instant, the hoist has an upward velocity of 2 ms and an acceleration of.5 ms 2. Determine the power generated by the motor M at this instant if it operates with an efficiency of 0.8. M Equations of Motion: Here,.5 ms 2. y referring to the free-body diagram of the hoist and counterweight shown in Fig. a, ; 2 800(9.8) 800(.5) ; Solving, N N Power: out 2T v 2( )(2) W Thus, () C in out e (0 3 ) W 9.5 kw

6 4 62. An athlete pushes against an exercise machine with a force that varies with time as shown in the first graph. Also, the velocity of the athlete s arm acting in the same direction as the force varies with time as shown in the second graph. Determine the power applied as a function of time and the work done in t = 0.3 s. F (N) t (s) For 0 t 0.2 v (m/s) F = 800 N 20 v = t = 66.67t P = F # v = 53.3 t kw 0.3 t (s) For 0.2 t 0.3 F = t v = 66.67t P = F # v = 60t - 533t 2 2 kw 0.3 u = Pdt L u = 53.3t dt + 60t - 533t 2 2 dt L L = (0.2) [(0.3)2 - (0.2) 2 ] [(0.3)3 - (0.2) 3 ] =.69 kj

7 4 68. Each of the two elastic rubber bands of the slingshot has an unstretched length of 80 mm. If they are pulled back to the position shown and released fromrest, determine the speed of the 30-g pellet just after the rubber bands become unstretched. Neglect the mass of the rubber bands. Each rubber band has a stiffness of 80 Nm. 240 mm 50 mm 50 mm (2) 2 (80)[(0.05)2 (0.240) 2 0.8] 2 2 (0.030)v2 v 4.76 ms

8 4 72. The 2-kg collar is attached to a spring that has an unstretched length of 3 m. If the collar is drawn to point and released from rest, determine its speed when it arrives at point A. 3 m k = 3 N/m Potential Energy: The initial and final elastic potential energy are and, respectively.the gravitational 2 (3)(3-3)2 = 0 2 (3)A = 6.00 J potential energy remains the same since the elevation of collar does not change when it moves from to A. A 4m Conservation of Energy: T + V = T A + V A = 2 (2) v2 A + 0 v A = 2.45 m s

9 4 89. When the 6-kg box reaches point A it has a speed of v A = 2m>s. Determine the angle u at which it leaves the smooth circular ramp and the distance s to where it falls into the cart. Neglect friction. 20 A v A = 2 m/s θ.2 m At point : s +b F n = ma n ; 6(9.8) cos f = 6a n2.2 b () Datum at bottom of curve: T A + V A = T + V 2 (6)(2)2 + 6(9.8)(.2 cos 20 ) = 2 (6)(v ) 2 + 6(9.8)(.2 cos f) = 0.5v cos f (2) Substitute Eq. () into Eq. (2), and solving for, v = 2.95 m>s Thus, f = cos - a (2.95)2.2(9.8) b = v u = f - 20 = 22.3 A + c s = s 0 + v 0 t + 2 a ct cos = (sin )t + 2 (-9.8)t t t = 0 Solving for the positive root: t = s a : + b s = s 0 + v 0 t s = 0 + (2.95 cos )(0.2687) s = m

10 4 92. The 75-kg man bungee jumps off the bridge at A with an initial downward speed of.5 m>s. Determine the required unstretched length of the elastic cord to which he is attached in order that he stops momentarily just above the surface of the water. The stiffness of the elastic cord is k = 80 N>m. Neglect the size of the man. A 50 m Potential Energy: With reference to the datum set at the surface of the water, the gravitational potential energy of the man at positions A and are AV g A = mgh A = 75(9.8)(50) = J and AV g = mgh = 75(9.8)(0) = 0. When the man is at position A, the elastic cord is unstretched (s A = 0), whereas the elastic cord stretches s = A50 - l 0 m, where is the unstretched length of the cord.thus, the l 0 elastic potential energy of the elastic cord when the man is at these two positions are AV and AV e =. 2 ks 2 = 2 (80)(50 - l 0) 2 = 40(50 - l 0 ) 2 e A = 2 ks A 2 = 0 Conservation of Energy: T A + V A = T + V 2 mv A 2 + av g b + AV e A R = A 2 mv 2 + av g b + AV e R 2 (75)(.52 ) + A = 0 + C0 + 40(50 - l 0 ) 2 D l 0 = 97.5 m