208_Physics ProblemsTechnical Physics

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online