Chapter 5 Work, Energy, Power Notes Answers Check Your Understanding 1 1. Ans (a) W.d = F.d = 50 J (b) W.d = !F cos30.s = 43.3!J (c) W.d = !F cos150.s = −43.3!J (d) W.d = !F cos90.s = 0 1
!2
2. By conservation of energy, m(v f 2 − vi 2 ) = mgΔh 1 2
(v −102 ) = 9.81(10 − 7)
2
−1
!v = 12.6!m!s
3. Force of 3-­‐kg mass along the slope = !mgsinθ = 3(9.81)sin30 Which is greater than weight of 1-­‐kg mass. Thus, 1-­‐kg mass will move upwards. Loss in GPE of 3-­‐kg mass = gain in GPE of 1-­‐kg mass and total gain in KE 1
3(9.81)(1sin30) = 1(9.81)(1)+ (3+1)v 2
2
−1
!v = 1.57!m!s
4. (a) loss in GPE = !mgh = 0.100(9.81)(20) = 19.62!J 1
!2
gain in KE = (0.100)152 = 11.25!J thus, w.d against drag = 19.62 – 11.25 = 8.37 J (b) W.d = Fave .s !
8.37
Fave =
= 0.42!N 20
!
5. by conservation of energy, loss in GPE = gain in EPE 1
mgΔh = kx 2 2
!
1
12(9.81)(3+1.3)sin30 = k(1.32 )
2
−1
!k = 279!N!m
Check Your Understanding 2 Δh
= mgv = (1800)(9.81)(5) = 88290!W (75%) t
!
88290
total power = ×100 = 1.18 ×105 !W ! 75
1. useful power = mg
2. (a) a =
!
Fnet
m
=
2000 − 800
= 3m!s −2 400
(b) !v = u + at = 0 + 3(5) = 15!m!s−1 1
2
1
2
KE = mv 2 = (400)(152 ) = 4.5×103 !J !
ΔE 4.5×103
=
= 9 ×103W t
5
!
(c) ave P = 3. At maximum speed, a = 0, F2!engine = Fdrag = kv = 10v !
P2!engines = F2!engines .v = k102 = 50!kW! × 2 !
!k = 1000 when one engine is working, P
= 50 ×103 !W = F1!engine .v = kv 2 = 1000v 2 ! 1!engine
−1
!v = 7.1!m!s 4. (a) !72!km!h−1 = 20!m!s−1 per second, distance moved along slope = 20 m 1
× 20 = 0.2!m !100
per second, gain in GPE = mgh = !2 ×105 (9.81)(0.2) = 3.92 ×105 !J per second, height gain = (b) per second, work done against friction = f.d = !1.28 ×104 × 20!m!=2.56 ×105 !J total power = !(3.92 + 2.56)×105 !J/s