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MECAˆNICA FUNDAMENTAL – 2013/2 – DF-ICEx FIS031 – Turma: B2 Respostas de problemas do Cap. 2 da Apostila Obs: Soluc¸a˜o do problema 2.3c corrigida 2.1 v = 3F0t0 m ;x(2t0) = 5 2 F0t20 m 2.2 a) v = F0t0 m ;x = F0t20 2m b) v = 1 m (F0t+ bt 2/2) ;x = 1 m (F0t 2/2 + bt3/6) c) v = F0 sen(ωt) mω ;x = F0(1−cos(ωt) mω2 d) v = kt 3 3m ;x = kt 4 12m 2.3 a) v = √ (2F0x+ kx2)m−1 b) v = √ 2F0 km (1− e−kx) c) x = m k2 [ kv − F0 ln ( F0+kv F0 )] 2.4 a) v = k x n+1 n+1 + C b) v = √ v20 − 2kxn+1m(n+1) c) x = ( m(n+1) 2k v20 )1/(n+1) 2.6 −v0 as + v0√ asad ; as = −g(sen θ + µ cos θ); ad = −g(sen θ − µ cos θ) 2.7 v = (A+Bt)α onde A = v1−n0 , B = c m (n− 1), α = (1− n)−1 x = C(vβ0 − vβ) onde C = mc (2− n)−1 e n 6= 1, 2 e β = 2− n 2.9 a) x = −m c |v + mg c ln(1 + cv mg )| b) x = −m 2c ln(1− cv2 mg ) 2.11 F (x) = −mb2x−3 2.13 x = a tg (bt) onde a = √ 2mv0/k, b = √ kv0/2m 2.14 T = 2piA v0 PA´GINA 1/2 2.15 A1 A2 √ m1 2m2 2.16 T = 2pi √ x22−x21 v21−v22 , A = √ x22v 2 1−x21v22 v21−v22 2.18 F = mg + mk m+M d cos (√ k m+M t ) , d = g (m+M) k 2.19 e2piγ/ω1 PA´GINA 2/2
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