Problem#1
The average speed of a nitrogen molecule in air is about 6.70 x 102 m/s, and its mass is 4.68 x 1026 kg. (a) If it takes 3.00 x 10-13 s for a nitrogen molecule to hit a wall and rebound with the same speed but moving in the opposite direction, what is the average acceleration of the molecule during this time interval? (b) What average force does the molecule exert on the wall?Answer:
(a) Let the x-axis be in the original direction of the molecule’s motion.
We use
vf = vi + at
–670 m/s = 670 m/s + a(3.00 x 10-13 s)
a = –4.47 x 1015m/s2
(b) For the molecule, ΣF = ma. Its weight is negligible.
Fwall on molecule = (4.68 x 1026 kg)(–4.47 x 1015m/s2) = –2.09 x 10-10 N
Fwall on wall = + 2.09 x 10-10 N
Problem#2
An electron of mass 9.11 x 1031 kg has an initial speed of 3.00 x 105 m/s. It travels in a straight line, and its speed increases to 7.00 x 105 m/s in a distance of 5.00 cm. Assuming its acceleration is constant, (a) determine the force exerted on the electron and (b) compare this force with the weight of the electron, which we neglected.
Answer:
Given:
m = 9.11 x 1031 kg
vi = 3.00 x 105 m/s
vf = 7.00 x 105 m/s
we use
vf2 = vi2 + 2a(xf – xi) and a = F/m, then
vf2 = 2(F/m)(xf – xi)
(7.00 x 105 m/s)2 = (3.00 x 105 m/s)2 + 2a(0.0500 m)
a = 3.99 x 1014 N
then
ΣF = ma = (9.11 x 10-31 kg)(3.99 x 1014 N) = 3.64 x 10-18 N
(b) The weight of the electron is
Fg = mg = (9.11 x 10-31 kg)(9.80 m/s2) = 8.93 x 10-30 N
The accelerating force is 4.08 × 1011 times the weight of the electron.
Problem#3
A woman weighs 120 lb. Determine (a) her weight in newtons (N) and (b) her mass in kilograms (kg).
Answer:
(a) Fg = mg = 120 lb = (4.448 N/lb)(120 lb) = 534 N
(b) m = Fg/g = 534 N/(9.80 m/s2) = 54.5 kg
Problem#4
If a man weighs 900 N on the Earth, what would he weigh on Jupiter, where the acceleration due to gravity is 25.9 m/s2?
Answer:
Fg = mg = 900 N,
m = Fg/g = 900 N/(9.80 m/s2) = 91.8 kg
(Fg)on Jupiter = 91.8 kg x (25.9 m/s2) = 2380 N
Problem#5
The distinction between mass and weight was discovered after Jean Richer transported pendulum clocks from Paris to French Guyana in 1671. He found that they ran slower there quite systematically. The effect was reversed when the clocks returned to Paris. How much weight would you personally lose in traveling from Paris, where g = 9.809 5 m/s2, to Cayenne, where g = 9.780 8 m/s2?
Answer:
Imagine a quick trip by jet, on which you do not visit the rest room and your perspiration is just canceled out by a glass of tomato juice. By subtraction,
(Fg)p = mgp and
(Fg)C = mgC give
ΔFg = m(gp – gC)
For a person whose mass is 88.7 kg, the change in weight is
ΔFg = (88.7 kg)(9.8095 – 9.7808) = 2.55 N
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