Solution to Pairs Exercise #1

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# Solution to Pairs Exercise #1 - PowerPoint PPT Presentation

Solution to Pairs Exercise #1. +. M=4000 kg. 20 m. +. +. Datum. Solution to Pairs Exercise #1. D E p =Net Mech. Work Input Net Mech. Work Input = D E p = mgh = (4000 kg)(9.8 m/s 2 )(20 m)(1 J·s 2 /(kg·m 2 )) = 784,000 J. Solution to Pairs Exercise #2.

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Solution to Pairs Exercise #1

+

M=4000 kg

20 m

+

+

Datum

Solution to Pairs Exercise #1

DEp=Net Mech. Work Input

Net Mech. Work Input = DEp

= mgh

= (4000 kg)(9.8 m/s2)(20 m)(1 J·s2/(kg·m2))

= 784,000 J

Solution to Pairs Exercise #2

h1= 0.35 (coal to electricity)

(see Example 21.24)

h2= 0.03 (Figure 21.30)

hoverall = h1 h2 = (0.35)(0.03)=0.0105

Solution to Bungee #1

Total Energy = Ep + Ek = mgh + ½ kx2 + ½ mv2

Energy on bridge = Energy at Xtaut

mg(0) + ½ k(0)2 + ½ m(0)2 = 75 kg (9.81 m/s2)(-50.0 m) + ½ k(0)2 + ½ (75 kg) v2

v = 31.3 m/s

Solution to Bungee #2

Total Energy = Ep + Ek = mgh + ½ kx2 + ½ mv2

Energy at bridge = Energy at hmax

mg(0) + ½ k(0)2 + ½ m(0)2 = (75 kg)(9.81 m/s2)(-hmax) +

½ (15 kg/s2)(hmax – 50 m)2 + ½ m(0)2

0 = -735.75 hmax + 7.5 (hmax)2 - 750 hmax + 18750

0 = 7.5 (hmax)2 – 1485.75 hmax + 18750

hmax = 184.6 m or 13.6 m

V = 0 at hmax

Stretched length only

Only valid solution

Solution to Bungee #3

Total Energy = Ep + Ek = mgh + ½ kx2 + ½ mv2

Energy at bridge = Energy at river

mmaxg(0) + ½ k(0)2 + ½ mmax (0)2 = mmaxg xriver +

½ k(xriver – xtaut)2 + ½ mmaxv2

0 = mmax (9.81 m/s2)(-195 m) + ½ (15 kg/s2)

(195 m – 50 m)2 + ½ mmax(0)2

mmax = 82.4 kg