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# 06.06 - 6.6 WORK 1 15 The force required to stretch the...

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Unformatted text preview: 6.6 WORK 1. 15. The force required to stretch the spring from its natural length of 2 m to a length of 5 In is F(x) = kx. The work done 3 3 byFisW = forced); :1: r0 xdx = \$112}: 2 %. This workisequalto 1800] => 31: = 1800 => k = 400N/m We ﬁnd the force constant from Hooke's law: F = kx. A force of 2 N stretches the spring to 0.02 m => 22kc(0.02) 22> km 100 Theforceof4Nwill stretchthe rubberbandym,wherekay 2:» ya: 4N 0.04 => y = W => y = 0.04m = 4cm. The workdone to stretch the rubber band 0.04misW = l; kxdx 0.04 0.04 =100f0 xdleool’gil =W§£323=0081 0 g _ 21.714 __ 21.714 x — 8—5 — 3 (a) We ﬁnd the springs constant from Hooke's law: F = kx => k 2 => k = 7238 g 0.5 0.5 0.5 (b) The work done to compress the assembly the first half inch is W = I; kx dx = 7238 I; x dx = 7238 0 = (7238) g = w x 905 in - lb. The work done to compress the assembly the second half inch is: 1.0 1.0 La W = M kx dx = 7238 r” x dx = 7238 = % [1 — (0.5)2] = (7238)““5’ m 2714111 - lb 0.5 2 The force required to haul up the rope is equal to the ropes weight, which varies steadily and is proportional to x, 50 50 the length of the rope still hanging: F(x) 2 0.624x. The work done is: W m I; F(x) dx 2 l; 0.624x dx 50 z 780] = 0.624 0 The force required to lift the cable is equal to the weight of the cable paid out: F(x) 2 (4.5)(180 — x) where x 80 is the position of the car off the first floor. The work done is: W = [:80 F(x) dx : 4.5 I: (180 — x) dx = 4.5 [mm — :80 = 4.5 (1002 — £8293) = 4551802 = 72,900 ft - lb We will use the coordinate system given. (a) The typical slab between the planes at y and y + Ay has a volume of AV 2 (10)(12) Ay = 120 Ay ft3. The force F required to lift the slab is equal to its weight: F = 62.4 AV : 62.4 - 120 Ay lb. The distance through which F must act is about y ft, so the work done lifting the slab is about AW = force x distance 2 62.4 - 120 - y - Ay ft - lb. The work it takes to lift all 20 the water is approximately W m 2 AW 0 20 = 2 62.4 - 120y - Ay ft - 1b. This is a Riemann sum for 0 the function 62.4 - 120y over the interval 0 5 y S 20. The work of pumping the tank empty is the limit of these sums: 20 2 20 w: [0 62.4-120ydy =(62.4)(120)[3’§l0 =(62.4)(120)(%) =(62.4)(120)(200)= 1,497,600ft-1b 17. 19. 21. The slab is a disk of area 17x2 = 2, thickness Ay, and height below the top of the tank (10 - y). So the work to pump the oil in this slab, AW, is 57(10 — y)7r(%)2. The work to pump all the oil to the top of the tank is 10 w = ﬂog-moﬁ — y3)dy = 22: [39\$ — £1] 0 = 11,8751r ft . 1b e 37,306ft . lb. The typical slab between the planes at y and and y + Ay has a volume of AV = 1r(radius)2(thickness) = 1r 2Ay = 1r - 100 Ay ft3. The force F required to lift the slab is equal to its weight: F = 51.2 AV 2 51.2 - 1007r Ay lb “2 F m 5120a Ay 1b. The distance through which F must act is about (30 — y) ft. The work it takes to lift all the 30 30 kerosene is approximately W m 2 AW = 2 512011130 — y) Ay ft - lb which is a Riemann sum. The work to pump the 0 0 39 30 tank dry is the limit of these sums: w = [0 51201r(30 — y) dy = 512071’ [30y — 3’51 z 7,238,229.48 ft - lb 0 = 5120111?) = (5120)(4501r) (a) Follow all the steps of Example 5 but make the substitution of 64.5 3% for 57 g. Then, [Isa—ﬁllw Les—E =(@)(83)(3—°-) 8 Wzl;@(lo_y)y2dy:% 4 0 4 3 4 = 64.51333 2 2157,. 33 m 34,582.65 ft - 1b (b) Exactly as done in Example 5 but change the distance through which F acts to distance a (13 — y) ft. Then 3 3 4 s D W: K,”T”<13-y>y2dy= 57+” l%"—--%lo= 577” @341 = (5%)(83H‘T3- = (197r) (82) (7)(2) R: 53.4825 ft - lb ___ 5777-83-7 2) _ 3-4 ...
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