A bag of sugar weighs 3.50 lb on Earth. What would it weigh in newtons on the Mo
ID: 1550282 • Letter: A
Question
A bag of sugar weighs 3.50 lb on Earth. What would it weigh in newtons on the Moon, where the free-fall acceleration is one-sixth that on Earth? N Repeat for Venus, where g is 0.904 times that on Earth. N Find the mass of the bag of sugar in kilograms at each of the three locations. Earth kg Moon kg Venus kg A freight train has a mass of 1.4 times 10^7 kg. If the locomotive can exert a constant pull of 8.3 times 10^5 N, how long does it take to increase the speed of the train from rest to 74 km/h? min A 77 kg man standing on a scale in an elevator notes that as the elevator rises, the scale reads 840 N. What is the acceleration of the elevator? m/s^2 upward A boat moves through the water with two forces acting on it. One is a 1, 800-N forward push by the water on the propeller, and the other is a 1, 400-N resistive force due to the water around the bow. (a) What is the acceleration of the 1, 100-kg boat? m/s^2 (b) If it starts from rest, how far will the boat move in 20.0 s? m (c) What will its velocity be at the end of that time? m/sExplanation / Answer
weight of sugar on moon W m = 3.5 (0.453592 kg) * (g/6)= 2.593 N
or
weight of sugar on moon W m = W e/ 6
= (3.5 / 6) lb
= 0.5833 lb
=0.5833 * 4.448 N
= 2.613 N
Weight of venus = mg = 3.50 (0.453592 kg) * 0.904g = 14.078N
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4) Given vi=0, vf=74km/hr=20.55m/s, m=1.4x107kg, F = 8.3x105N.
Choose equations. F=ma, and vf=vi+at.
Solve equations. a=F/m,
t=(vf-vi)/a
t= (vf-vi)m/F
t= (20.55m/s)(1.4x107kg)/8.3x105N = 346.62 kg m/sN =346.62 s or about 5.777 min
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5.
F = mg + ma
ma = F - mg
a = ( F - m g ) / m
= (840 N - (77 kg ) (9.8 m/s^2 ) ) / 77 kg
= 1.109 m/s
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6.
Ftotal = Fresistive + Faccelerating
1800 N = 1400 + Faccelerating
Faccelerating = 1800N - 1400N = 400 N
Use F=ma, solved for a, to determine the acceleration
a = F/m = (400)/(1000 kg) = 0.4 m/s2
(b) Determine the distance it will travel using x = vit +(1/2)at2
x = vit +(1/2)at2 = (0)(20.0s) + (1/2)(0.4 m/s^2)(20.0 s)2
x = (1/2)(0.400 m/s2)(400. s2) = (0.1 m)(100.) = 80m
(c) Determine the final velocity using vf = vi + at
vf = vi + at = (0 m/s) + (0.4 m/s2)(20.0 s) = 8 m/s
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