# Solucionario van wylen

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2-1

CHAPTER 2
The correspondence between the problem set in this fifth edition versus the problem set in the 4'th edition text. Problems that are new are marked new and those that are only slightly altered are marked as modified (mod). New 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Old 4 mod new new 7 mod 2 mod new new new 5 mod 6 8 mod new 9 mod 10 mod 11 new new 16 mod new 12 New 2122 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39E 40E Old 13 14 15 17 18 new 19 20 21 22 23 24 new 25 mod 26 mod 27 mod 28 29 31E mod 32E New 41E 42E 43E 44E 45E 46E 47E 48E 49E Old 33E mod 34E mod 35E 36E 37E 38E 39E 40E 41E

2-2 2.1 The “standard” acceleration (at sea level and 45° latitude) due to gravity is 9.80665 m/s2. What is the force needed to hold a mass of 2 kg at rest in thisgravitational field ? How much mass can a force of 1 N support ? Solution: ma = 0 = ∑ F = F - mg F = mg = 2 × 9.80665 = 19.613 N F = mg 2.2 => m = F/g = 1 / 9.80665 = 0.102 kg

A model car rolls down an incline with a slope so the gravitational “pull” in the direction of motion is one third of the standard gravitational force (see Problem 2.1). If the car has a mass of 0.45 kg. Find theacceleration. Solution: ma = ∑ F = mg / 3 a = mg / 3m = g/3 = 9.80665 / 3 = 3.27 m/s2

2.3

A car drives at 60 km/h and is brought to a full stop with constant deceleration in 5 seconds. If the total car and driver mass is 1075 kg. Find the necessary force. Solution: Acceleration is the time rate of change of velocity. ma = ∑ F ; a = dV / dt = (60 × 1000) / (3600 × 5) = 3.33 m/s2 Fnet = ma = 1075 ×3.333 = 3583 N

2.4

A washing machine has 2 kg of clothes spinning at a rate that generates an acceleration of 24 m/s2. What is the force needed to hold the clothes? Solution: F = ma = 2 kg × 24 m/s2 = 48 N

2.5

A 1200-kg car moving at 20 km/h is accelerated at a constant rate of 4 m/s2 up to a speed of 75 km/h. What are the force and total time required? Solution: a = dV / dt => ∆t = dV/a= [ ( 75 − 20 ) / 4 ] × ( 1000 / 3600 ) ∆t = 3.82 sec ; F = ma = 1200 × 4 = 4800 N

2-3 2.6 A steel plate of 950 kg accelerates from rest with 3 m/s2 for a period of 10s. What force is needed and what is the final velocity? Solution: Constant acceleration can be integrated to get velocity. a = dV / dt => V = 30 m/s ; 2.7

∫ dV = ∫ a dt

=> ∆V = a ∆t = 3 × 10 = 30 m/s

F = ma = 950 × 3 =2850 N

A 15 kg steel container has 1.75 kilomoles of liquid propane inside. A force of 2 kN now accelerates this system. What is the acceleration? Solution: ma = ∑ F ⇒ a = ∑ F / m m = msteel + mpropane = 15 + (1.75 × 44.094) = 92.165 kg a = 2000 / 92.165 = 21.7 m/s2

2.8

A rope hangs over a pulley with the two equally long ends down. On one end you attach a mass of 5 kg and on the otherend you attach 10 kg. Assuming standard gravitation and no friction in the pulley what is the acceleration of the 10 kg mass when released? Solution: Do the equation of motion for the mass m2 along the downwards direction, in that case the mass m 1 moves up (i.e. has -a for the acceleration) m2 a = m2 g − m1 g − m1a (m1 + m2 ) a = (m2 − m1 )g This is net force in motion direction a = (10 − 5) g /(10 + 5) = g / 3 = 3.27 m/s2

g

2 1

2.9

A bucket of concrete of total mass 200 kg is raised by a crane with an acceleration of 2 m/s2 relative to the ground at a location where the local gravitational acceleration is 9.5 m/s2. Find the required force. Solution: F = ma = Fup − mg Fup = ma + mg = 200 ( 2 + 9.5 ) = 2300 N

2-4 2.10 On the moon the gravitational acceleration isapproximately one-sixth that on the surface of the earth. A 5-kg mass is “weighed” with a beam balance on the surface on the moon. What is the expected reading? If this mass is weighed with a spring scale that reads correctly for standard gravity on earth (see Problem 2.1), what is the reading? Solution: Moon gravitation is: g = gearth/6

m

m
Beam Balance Reading is 5 kg This is mass comparison...