Homework 6 Solutions
a)W = F · dW = 720[J ] , d = 2.2[m]
F =W
d
F =720[J ]
2.2[m]
F = 327[N ]
b) m2 = 6m , F2 = F , d2 = 2dW2 = F2 · d2W2 = F · 2d = 2(F · d)W2 = 2W2 times more work
1
a)The displacement is horizontal =⇒ only horizontal forces will do work
(not gravity or normal force)
The pull form the string and friction form the floor
b) Because the displacement is in the horizontal direction, only the hori-zontal component Fh will do work
a) Fh = 76[N ] , Fv = 27[N ]
Sled is not moving vertically =⇒ W=0[J]
b)W = F · dW = 76[N ] · 26[m]
W = 1980[J ]
c) Wtotal = Wh + Wv
Wtotal = 1976[J ] + 0[J ]
Wtotal = 1976[J ]
a) m = 6.6[kg] , h = 4.0[m]P.E. = mghP.E. = 6.6[kg]9.8[m
s2]4.0[m]
P.E. = 259[J ]
b) W = ∆E = Ef − Ei
W = P.E.f − P.E.i
W = 259[J ]
c) P =W
t
P =259[J ]
2.0[s]
P = 129[W ]
a) m = 145[g] = 0.145[kg] , v = 63[mihr ]1609[m]
1[mi]1[hr]3600[s] = 28.16[ms ]
K.E. = 12 mv2
K.E. = 12 0.145[kg](28.16[ms ])2
K.E. = 57.5[J ]
b) W = ∆E = Ef − E − iW = K.E.f −K.E.i
|W | = 57.5[J ]
c) d = 2.5[cm] = 0.025[m]
F = Wd
F = 57.5[J ]0.025[m]
F = 2300[J ]
d) m2 = m , v2 = 15v
K.E.2 = 12 m2v
22
K.E.2 = 12 m(15v)2
K.E.2 = 12 mv2
25 = 125(12 mv2)
K.E.2 = 125K.E.
1
25= 0.04
a)If the force is perpendicular to the direction of motion, then the work is
0. Also, if the force is in the opposite direction of motion, it will slow theobject down and decrease the kinetic energy. Finally, only if the force is inthe direction of motion, it would increase the kinetic energy.
b)The total energy of the universe is always conserved. This is because
there are no outside forces acting on the universe to do work. Mechanicalenergy is conserved only when springs and gravity do work on the system.If there was another force, it would do work on the system and not storeenergy.
a)the pendulum isn’t at its highest point or moving the fastest, so it has a
combination of kinetic energy and potential energy.
b)The work done on the pendulum is form the outside force pulling it back
from equilibrium =⇒ work done gives the initial potential energy.
c)At the highest point, all the energy is potential energy, because it is not
moving anymore (K.E. = 0).
d)At the lowest point, all the potential energy has been converted into ki-
netic energy.
a) m = 2.77[kg] , slowest point = 13[ms ]
K.E. = 12 mv2
K.E. = 12 2.77[kg](13[ms ])2
K.E. = 234[J ]
b) W = ∆E = Ef − E − i = 0 =⇒Ef = Ei
P.E. = K.E.mgh = K.E.
h =K.E.
mg
h =234[J ]
2.77[kg]9.8[ms2
]
h = 8.62[m]
c) 5 oscillations in 8.2[s]T = time for one oscillation
T =8.2[s]
5[oscillations]
T = 1.64[s]
d) f = 1T
f =1
1.64[s]
f = 0.61[Hz]
m = 5.0[kg] , P.E.i = 175[J ] , P.E.f = 90[J ]W = ∆E = Ef − Ei = 0 =⇒Ei = Ef
P.E.i + K.E.i = P.E.f + K.E.fK.E.f = P.E.i + K.E.i − P.E.fK.E.f = 175[J ] + 0[J ]− 90[J ]
K.E.f = 85[J ]
a) k = 680[Nm ] , m = 40[g] = 0.040[kg] , d = 38[cm] = 0.38[m]
E.P.E. = 12 kx2
E.P.E. = 12 680[Nm ](0.38[m])2
E.P.E. = 49.1[J ]
b) W = ∆E = Ef − Ei
W = E.P.E.f − E.P.E.iW = 49.1[J ]− 0[J ]
W = 49.1[J ]
c) W = ∆E = Ef − Ei = 0Ei = Ef
E.P.E. = K.E.K.E. = 49.1[J ]
d) K.E. = 12 mv2
v =
√2 ·K.E.
m
v =
√2 · 49.1[J ]
0.040[kg]
v = 49.5[m
s]
e) W = ∆E = Ef − Ei = 0Ei = Ef
K.E. = P.E.K.E. = mgh
h =K.E.
mg
h =49.1[J ]
0.040[kg]9.8[ms2
]
h = 125[m]
P.E.i = 1.80[J ] , P.E.f = 0[J ] , K.E.i = 0[J ] , K.E.f = 1.11[J ]W = ∆E = Ef − Ei
W = P.E.f + K.E.f − (P.E.i + K.E.i)W = 0[J ] + 1.11[J ]− 1.80[J ]− 0[J ]
W = −0.69[J ]
a) m = 3900[kg] , h = 33[m] , vtop = 15[ms ]P.E. = mghP.E. = 3900[kg]9.8[ma ]33[m]
P.E. = 1.26E6[J ]
b) K.E. = 12 mv2
K.E. = 12 3900[J ](15[ms ])2
K.E. = 4.39E5[J ]
c) Emech = P.E. + K.E.Emech = 1.26E6[J ] + 4.39E5[J ]
Emech = 1.70E6[J ]
d) hf = 13[m]W = ∆E = Ef − Ei = 0 =⇒Ei = Ef
Emech = K.E.f + P.E.fK.E.f = Emech −mghfK.E.f = 1.70E6[J ]− 3900[kg]9.8[m
s2]13[m]
K.E.f = 1.20E6[J ]
e) W = ∆E = Ef − Ei
W = P.E.f + K.E.f − Emech
W = mghf + K.E.f − Emech
K.E.f = W −mghf + Emech
K.E.f = −4.29E5[J ]− 3900[kg]9.8[ms2
]13[m]− 1.70E6[J ]K.E.f = 7.74E5[J ]
a) d = 8.10[cm] = 0.0810[m] , F = 40.0[N ]F = −kxk = |F |
x
k = 40.0[N ]0.0810[m]
k = 494
[N
m
]b) E.P.E. = 1
2 kx2
E.P.E. = 12 494[Nm ](0.0810[Nm ])2
E.P.E. = 1.62[J ]
