PEi KEi Wc PEf KEf Wn 1 BEFORE
PEi + KEi + Wc = PEf + KEf + Wn 1) BEFORE AFTER KEf = 0 PEf = mgh Wc = PEf Wc = mgh Substitute: Wc = 100(9. 8)(1. 5) 1. 5 m KEi = 0 PEi = 0 W = ? c Wn = 0 Wc = 1470 J 2) BEFORE AFTER PEi + KEi + Wc = PEf + KEf + Wn 3 (sin 30) = h h = 1. 5 m KEi = 0 PEi = mgh h Wc = 0 Wn = 0 3. 0 m ½mv 2 KEf = PEf = 0 PEi = KEf mgh = ½mv 2 Substitute: (2)(9. 8)(1. 5) = ½(2)(v 2) v = 5. 4 m/s
3) BEFORE AFTER PEi + KEi + Wc = PEf + KEf + Wn KEi = Wn ½mv 2 KEi = PEi = 0 KEf = 0 PEf = 0 ½mv 2 = fΔd Substitute: ½(m)(8)2 = (3. 92 m)(Δd) Wc = 0 Wn = fΔd f =μFN FN = Fg = 9. 8 m f =(0. 4)(9. 8 m) = 3. 92 m Δd = 8. 16 m
4) BEFORE AFTER PEi + KEi + Wc = PEf + KEf + Wn Wc = PEf KEi = 0 PEi = 0 KEf = 0 PEf = mgh Wc = mgh Substitute: Wc = (3. 0)(9. 8)(0. 15) Wc = ? Wn = 0 Wc = 4. 41 J 1. 30° h 1. 25 m Need to use trig to find h adj cos(30) = 1. 25 adj = 1. 1 m Now use ‘after’ diagram as ‘before’ and make an ‘after’ diagram – then solve for velocity at the bottom v = 1. 7 m/s thus h = 1. 25 – 1. 1 = 0. 15 m
5) BEFORE AFTER KEi = 0 PEi = ½kx 2 KEf = ½mv 2 PEf = mgh PEi + KEi + Wc = PEf + KEf + Wn PEi = PEf + KEf ½kx 2 = mgh + ½mv 2 Substitute: ½ 800(0. 04)2 = (0. 025)(9. 8)(0. 15) + ½(0. 025)v 2 Wc = 0 Wn = 0 Now use ‘after’ diagram as ‘before’ and make an ‘after’ diagram – then solve for maximum height V = 6. 9 m/s h = 2. 45 m above the muzzle (2. 6 m above the initial position)
6) BEFORE AFTER PEi + KEi + Wc = PEf + KEf + Wn PEi = KEf + Wn KEi = 0 PEi = mgh h Wc = 0 Wn = fΔd mgh = ½mv 2 + fΔd 3. 0 m KEf = ½mv 2 PEf = 0 Need to use trig to find h 3 (sin 30) = h h = 1. 5 m Substitute: f =μFN (2)(9. 8)(1. 5) = ½(2)(v 2) + (3. 4)(3. 0) FN = Fg cos θ = (2)(9. 8)(cos 30) = 17 N f =(0. 2)(17) = 3. 4 N v = 4. 4 m/s
7) BEFORE AFTER KEf = ½m 1 v 2 PEf = 0 KEf = ½m 2 v 2 PEf = 0 KEi = 0 PEi = m 2 gh Substitute: PEi = 0 Wc = 0 (3)(9. 8)(. 8) = ½ (5)(1. 5)2 + ½(3)(1. 5)2 + (f)(. 8) Wn = fΔd f = 18. 15 N Equation: f =μFN PE + KE + W = PE + KE + W i i c f f PEi = KEf + Wn m 2 gh = ½m 1 v 2 + ½m 2 v 2 + fΔd n FN = Fg = (5)(9. 8) = 49 N 18. 15 = μ(49) μ = 0. 37
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