This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 62.4120ydy =(62.4)(120)[3’§l0 =(62.4)(120)(%) =(62.4)(120)(200)= 1,497,600ft1b 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 = ﬂogmoﬁ — 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”<13y>y2dy= 57+” l%"—%lo= 577” @341 = (5%)(83H‘T3 = (197r) (82) (7)(2) R: 53.4825 ft  lb ___ 5777837
2) _ 34 ...
View
Full Document
 Spring '09
 Cal
 Force, Work, Ay

Click to edit the document details