Problem#1
A 0.400-kg particle slides around a horizontal track. The track has a smooth vertical outer wall forming a circle with a radius of 1.50 m. The particle is given an initial speed of 8.00 m/s. After one revolution, its speed has dropped to 6.00 m/s because of friction with the rough floor of the track. (a) Find the energy converted from mechanical to internal in the system due to friction in one revolution. (b) Calculate the coefficient of kinetic friction. (c) What is the total number of revolutions the particle makes before stopping?Answer:
(a) The energy converted from mechanical to internal in the system due to friction in one revolution. We use
∑W = ∆K
Wother + (–∆Eint) = ∆K
0 –∆Eint = ∆K
∆Eint = – (½ mvf2 – ½ mvi2) = – ½ (0.400 kg)[(6.00 m/s)2 - (8.00 m/s)2] = 5.60 J
(b) the coefficient of kinetic friction is
∆Eint = fk∆r = µkmg(2πr)
5.60 J = µk(0.400 kg)(9.80 m/s2)2π(1.50 m)
µk = 0.152
(c) the total number of revolutions the particle makes before stopping.
After N revolutions, the object comes to rest and Kf = 0.
Wother + (–∆Eint) = ∆K
0 + ∆Eint = 0 + ½ mvi2
µkmg(N2πr) = ½ mvi2
This gives,
N = vi2/(4πµkgr)
N = (8.00 m/s)2/[(4π)(0.152)(9.80 m/s2)(1.50 m)]
N = 2.28 rev
Problem#2
In diatomic molecules, the constituent atoms exert attractive forces on each other at large distances and repulsive forces at short distances. For many molecules, the Lennard-Jones law is a good approximation to the magnitude of these forces:
F = F0[2(σ/r)13 – (σ/r)7]
where r is the center-to-center distance between the atoms in the molecule, σ is a length parameter, and F0 is the force when r = σ. For an oxygen molecule, we find that F0 = 9.60 x 10-11 N and σ = 3.50 x 10-10 m. Determine the work done by this force if the atoms are pulled apart from r = 4.00 x 10-10 m to r = 9.00 x 10-10 m.
Answer:
If positive F represents an outward force, (same as direction as r), then
W = ∫F.dr
= ∫rirf F0[2(σ/r)13 – (σ/r)7]dr
= ∫rirf (2F0σ13r-13 – F0σ7r-7)dr
= (–F0σ13r-12/6 + F0σ7r-6/6)│rirf
= –F0σ13(rf-12 – ri-12)/6 + F0σ7(rf-6 – ri-6)/6
= –(9.60 x 10-11 N)(3.50 x 10-10 m)13(rf-12 – ri-12)/6 + (9.60 x 10-11 N)(3.50 x 10-10 m)7(rf-6 – ri-6)/6
= –1.89 x 10-134(3.54 x 10-12 – 5.96 x 10-8) x 10120 + 1.03 x 10-77(1.88 x 10-6 – 2.44 x 10-6) x 1060
W = –1.37 x 10-21 J
Problem#3
As it plows a parking lot, a snowplow pushes an evergrowing pile of snow in front of it. Suppose a car moving through the air is similarly modeled as a cylinder pushing a growing plug of air in front of it. The originally stationary air is set into motion at the constant speed v of the cylinder, as in Figure 1. In a time interval ∆t, a new disk of air of mass ∆m must be moved a distance v∆t and hence
must be given a kinetic energy ½ (∆m)v2. Using this model, show that the automobile’s power loss due to air resistance is ½ ρAv3 and that the resistive force acting on the car is ½ ρAv2, where ρ is the density of air. Compare this result with the empirical expression ½ DρAv2 for the resistive force.
we know that
W = ∆K
Pt = ½ ∆mv2
The density is
ρ = ∆m/vol = ∆m/(A∆x)
∆m = ρA∆x
Substituting this into the first equation and solving for P,
Pt = ½(ρA∆x)v2
P = ½(ρA∆x/t)v2
Since, ∆x/∆t = v, then
P = ½ρAv3
Also, since P = Fv, so that
F = ½ρAv2
Our model predicts the same proportionalities as the empirical equation, and gives D = 1 for the
drag coefficient. Air actually slips around the moving object, instead of accumulating in front of it.
For this reason, the drag coefficient is not necessarily unity. It is typically less than one for a
streamlined object and can be greater than one if the airflow around the object is complicated.
Problem#4
A particle moves along the x axis from x = 12.8 m to x = 23.7 m under the influence of a force
F = 375/(x3 + 3.75x)
where F is in newtons and x is in meters. Using numerical integration, determine the total work done by this force on the particle during this displacement. Your result should be accurate to within 2%.
Answer:
We evaluate
W = ∫Fdx
W = ∫12.823.7[375/(x3 + 3.75x)]
by calculating
375(0.100)/[(12.8)3 + 3.75(12.8)] + 375(0.100)/[(12.9)3 + 3.75(12.9)] + . . . + 375(0.100)/[(23.6)3 + 3.75(23.6)] = 0.806
and
375(0.100)/[(12.9)3 + 3.75(12.9)] + 375(0.100)/[(13.0)3 + 3.75(13.0)] + . . . + 375(0.100)/[(23.7)3 + 3.75(23.7)] = 0.791
The answer must be between these two values. We may find it more precisely by using a value for
∆x smaller than 0.100. Thus, we find the integral to be 0.799 Nm.
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