( ) = 0.43 kj = 430 J. Solutions 9 1. Solutions to Miscellaneous Exercise 9 1. Let W = work done then 0.

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1 Soluions 9 Soluions o Miscellaneous Exercise 9. Le W work done hen.9 W PdV Using Simpson's rule (9.) we have. W { 96 + [ ] + [ ]+ }. kj. Using Simpson's rule (9.) again: W [ ]+.55+ [. +.88]+.6. kj J. { }. In his case we canno use Simpson's rule because we need an even number of inervals so we apply he rapezium rule. 5 Δh.57 + ( ) kj / kg ( s.f.). Esablishing a able ln Using Simpson's rule:.5 h + ( ) + ( 8.5 ) m ( s.f.) 5. Le's apply he Simpson's rule wih widh h. v v. ( ) ( ) + (.55 ) +.58.s ( s.f.) 6. Apply Simpson's rule wih h.: sin()/ By (9.7) he mean value, M, is given by sin () M d... {.97 ( ) (.9.9 ).8 }.9 V using Simpson h b (9.) y { y + ( y+ y +...) + ( y + y +...) + yn} a b (9.7) Mean value of y y b a a

2 Soluions 9 7. We use numerical inegraion o find he mean value of i over he inerval o. Use Simpson's rule wih h 8: i sin() Applying (9.7) gives Mean value of i sin() d 8. We have Area - 9 x x - 8 (..69 ) (.55 ) [ 7.9 ].9A x x x x sin By (8.), u and a + 6 () ( ) + sin ( ) + sin 6 6 sin sin Simplifyi ( ng) [ ] (Since he funcion is even we could inegrae beween and and hen double our resul). 9. Using (9.7) (8.) Mean value of v 5 ( e ) ( e ) d (9.7) Mean value of d 5 Because e ( Inegraing) 5 ( e ) ( e ).V ( d.p. ) u a u a u du a u + sin a b y y b a a

3 Soluions 9 Le v R. M.S represen he R.M.S. value of v : ( vrm.. S) 5 ( e ) d ( e + e ) d Because 5 e e ( Inegraing) ( + 8e e ) ( + 8e e ) V To find he R.M.S. value we ake he square roo: v RM.. S V ( s.f.). (a) Inegraing gives v d d + 9.8d vln ( ) C Subsiuing, v ln ( ) + C gives C ln ( ) Thus we have v ln ln ( ) ( ) ( ) ( ) ( ) ln ln Taking Ou v ln (b) We use numerical inegraion for inegraing v o find he heigh. Using Simpson's rule wih inerval 5 6 v ln [ ] heigh ( )+.8+ (.) m ( s.f.). We have P 96V. Work done V dv 96 V dv.9. V J 5.5 kj s.f. ( )

4 Soluions 9. Using (9.) we have d y.5 ( x 5 ) 8 ( x ) 5 x +.5 ( x x 5 ) 8 x x.5 x 5 x x d y.5 x x Inegraing wice: dy.5x x 8.5x C + +.5x x 8.5x y C x D Subsiuing x, y ino (*) [ ] + C( ) + D gives D Subsiuing x 5, y ino (*) C + + [ 5.5 ] + 5 C C 5 Puing C.65 and D ino (*).5x x 8.5x y.65x +. (i) We have d x d Inegraing gives C where C is a consan. d Using, d v cos () θ gives C v cos ( θ ) v cos( θ ) d Inegraing again wih respec o gives x vcos( θ ) + D where D is a consan (*) (9.) d y M

5 Soluions 9 5 Using, x gives D. Hence x vcos( θ ) (ii) We have d y g d Inegraing gives dy g k k inegraion consan d + Using dy, d v sin ( θ ) gives k vsin () θ We have dy g + v sin θ d ( ) Inegraing again gives g y + vsin ( θ ) + E Using, y gives E. Hence g y vsin ( θ ) 5. Subsiuing a, b ω ino (9.8) gives ω ω vrm.. S. v d ( ) ( ) To show he resul in he quesion, we need only evaluae subsiue our answer ino (). ( ω ) ( ω ) sin + sin v V V ( ) ( ω ) ( ω ) ( ω ) ( ) ( ω ) ( ω ) ( ω ) ( ω ) () and separaing he inegrals gives ω vd Vsin + VV sin sin + Vsin ω v V sin + V sin + VV sin sin Subsiuing his ino ω ω ( ) ( ) ω V sin ω d+ V sin ω d ( v RM.. S. ) ω + VV sin ( ω) sin( ω) d How do we evaluae he firs inegral in ( )? We can use (.68): ω ω sin ( ω) d cos( ω) d and ( ) (9.8) ( ) v RM.. S. b v b a d a (.68) sin ( A) cos( A)

6 Soluions 9 6 ω cos ( ω) sin ω ( ω) sin ω ω ω ω ω ω Similarly sin ( ω) d. ω How do we inegrae sin ω sin ω ω d ( ) ( ) wih respec o? Use (.59) and hen inegrae he resul. ω ω ( ω ) ( ω ) ( ω ω ) ( ω + ω ) sin sin d cos cos d ω ( ω ) ( ω ) cos cos d ( ω) ( ω) sin sin ω ω sin ω sin ω ω ω ω ω ω ω sin ω d, sin ω d ω and ω sin( ω)sin( ω)d ino ( ) gives Puing ( ) ( ) ω ω ( vrm.. S. ) V V VV ( ) + + ω ω ω V + V V + V ω Thus he resul required. ω (.59) sin( A) sin( B) cos( AB) cos( A+ B)