208_Physics ProblemsTechnical Physics

208_Physics ProblemsTechnical Physics - 210 P7.60 Energy...

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Unformatted text preview: 210 P7.60 Energy and Energy Transfer (a) a fe j e20.5i + 14.3 jj N = a 42.0 N fecos 150° i + sin 150° jj = e−36.4i + 21.0 jj N F1 = 25.0 N cos 35.0° i + sin 35.0° j = F2 e−15.9 i + 35.3 jj N (b) ∑ F = F1 + F2 = (c) a= (d) v f = v i + at = 4.00 i + 2.50 j m s + −3.18 i + 7.07 j m s 2 3.00 s ∑F = e−3.18 i + 7.07 jj m s e vf = (e) m 2 j e ja je e−5.54i + 23.7 jj m s r f = ri + v i t + e 12 at 2 f 1 e−3.18i + 7.07 jjem s ja3.00 sf 2 jb ga e−2.30i + 39.3 jj m 2 r f = 0 + 4.00 i + 2.50 j m s 3.00 s + ∆r = r f = (f) Kf = (g) b 1 1 mv 2 = 5.00 kg f 2 2 g a5.54f + a23.7f em s j = 2 2 2 2 Kf = 1 mvi2 + ∑ F ⋅ ∆r 2 1 2 K f = 5.00 kg 4.00 + 2.50 2 K f = 55.6 J + 1 426 J = 1.48 kJ b P7.61 f (a) 1.48 kJ g a f a f bm sg + a−15.9 Nfa−2.30 mf + a35.3 Nfa39.3 mf ∑ W = ∆K : 2 2 Ws + W g = 0 a f ja fa 12 kxi − 0 + mg∆x cos 90°+60° = 0 2 1 2 1.40 × 10 3 N m × 0.100 − 0.200 9.80 sin 60.0° ∆x = 0 2 ∆x = 4.12 m e (b) ∑ W = ∆K + ∆Eint : fa fa f Ws + W g − ∆Eint = 0 12 kxi + mg∆x cos 150°− µ k mg cos 60° ∆x = 0 2 1 2 1.40 × 10 3 N m × 0.100 − 0.200 9.80 sin 60.0° ∆x − 0.200 9.80 0.400 cos 60.0° ∆x = 0 2 ∆x = 3.35 m e ja fa fa fa f a fa fa fa f ...
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