Homework 8 solutions
Notes: Please keep in mind that most of the problems in MasteringPhysics with numerical answers are randomized for each student. The solutions below correspond to the version of the problem that is in the paper copy of the textbook. There may be some differences between the textbook version and the online version.
1. Reading questions
a. As the person ascends the stairs, some internal energy of the person is converted into an increase in the gravitational potential energy of the person+Earth system. Although this increase is the same regardless of whether the person walked or ran up the stairs, the rate at which the internal energy converted to potential was higher when the person ran. This is why it is harder to run up the stairs compared with walking.
b. The work-energy equation is only helpful in collisions when very little of the kinetic energy of the system is converted into internal energy because this conversion is hard to quantify. The impulse-momentum equation is not sensitive to conversions such as this and as a result is more useful in collisions where this sort of thing occurs (an example is a collision between two cars where the cars suffer significant damage).
2. Tutorial (the problem parts plus the hints in MasteringPhysics walk you through how to do it)
3. Tutorial (the problem parts plus the hints in MasteringPhysics walk you through how to do it)
4. Problem 6.29
Sketch and translate A sketch of the process is shown below. The system is the car, the road surface, and Earth. The origin of a vertical y-axis will be at the bottom of the hill with the positive y-axis pointing up. The car rolls 50 m down a 5.0º incline surface to reach the bottom. The internal energy of the system has increased due to friction.
Simplify and diagram We represent the process with a bar chart, shown above right.
Represent mathematically By energy conservation principle, we have
Ugi+Ki =Ugf +Kf +ΔUint ⇒ mgh+0=0+1 2mvf
2 + fl
Solve and evaluate The initial total energy of the system is
Ugi =mgh=mglsinθ=
(
900 kg)
(
9.8 m/s2)
(
50 m)
sin5.0° =3.84×104 Jvf = 2gh−2fl
m = 2 9.8 m/s 2
(
)
(
50 m)
sin5.0°−2 400 N(
)
(
50 m)
900 kg =6.4 m/s At the bottom, flat surface, the distance the car travels before coming to a stop is given by fd=1
2mvf 2
, or
d= mvf
2
2f =
900 kg
(
)
(
6.4 m/s)
22 400 N
(
)
=46 mThe general expression for the stopping distance is
d = mvf 2
2f = m
2f 2gh− 2fl
m ⎛
⎝⎜
⎞ ⎠⎟=
mgh f −l
Thus, we see that the stopping distance is greater for a more massive car, under the same force of friction f. If the mass of the car is 1800 kg instead, then
d′= m′gh
f −l=
1800 kg
(
)
(
9.8 m/s2)
(
50 m)
sin5.0°400 N −50 m=142 m
5. Problem 6.63
(a) Let the barbell be lifted up by 1.0 m, about the arm length. The work done is equal to the increase in gravitational potential energy:
W =ΔUg =mgΔy=
(
5.5 kg)
(
9.8 m/s2)
(
1.0 m)
≈54 J(b) Suppose it takes 0.3 s to lift the barbell up, the mechanical power is then
P= ΔW
Δt
= 54 J
0.30 s
=180 W
(c) With ΔU
int =300 kcal=(300 kcal)(4186 J/kcal)=1.26×10 6
J, at 20% efficiency, the number of times one needs to lift the barbell to burn off the energy is
N =εΔNint
ΔU g
= (0.20)(1.26×10
6 J)
54 J ≈4.7×10
