14–18.

The two blocks A and B have weights andIf the kinetic coefficient of friction between the

incline and block A is determine the speed of Aafter it moves 3 ft down the plane starting from rest. Neglectthe mass of the cord and pulleys.

mk = 0.2,WB = 10 lb.

WA = 60 lb

SOLUTIONKinematics: The speed of the block A and B can be related by using positioncoordinate equation.

(1)

Equation of Motion: Applying Eq. 13–7, we have

Principle of Work and Energy: By considering the whole system, which acts inthe direction of the displacement does positive work. and the friction force

does negative work since they act in the oppositedirection to that of displacement Here, is being displaced vertically (downward)

and is being displaced vertically (upward) . Since blocks A and B areat rest initially, . Applying Eq. 14–7, we have

(2)

Eqs. (1) and (2) yields

Ans.

vB = 7.033 ft s

vA = 3.52 ft>s

1236.48 = 60v2A + 10v2

B

60B 35

(3)R - 9.60(3) - 10(6) =12¢ 60

32.2≤v2

A +12¢ 10

32.2≤v2

B

0 + WA¢35

¢sA≤ - Ff¢sA - WB¢sB =12

mAv2A +

12

mB v2B

T1 + aU1 - 2 = T2

T1 = 0¢sBWB

35

¢sA

WA

Ff = mkN = 0.2(48.0) = 9.60 lbWB

WA

+ ©Fy¿ = may¿ ; N - 60a45b =

6032.2

(0) N = 48.0 lb

2vA - vB = 0

2¢sA - ¢sB = 0 ¢sB = 2¢sA = 2(3) = 6 ft

sA + (sA - sB) = l 2sA - sB = l

BA

5

4

3

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*14–40.

Cylinder A has a mass of 3 kg and cylinder B has a mass of8 kg. Determine the speed of A after it moves upwards 2 mstarting from rest. Neglect the mass of the cord and pulleys.

SOLUTION

Also, , and because the pulleys are massless, . The and terms drop out and the work-energy equation reduces to

Ans.vA = 3.82 m>s255.06 = 17.5v2

A

F2F1F1 = 2F2vB = 2vA

0 + 2[F1 - 3(9.81)] + 4[8(9.81) - F2] =12

(3)v2A +

12

(8)v2B

a T1 + a U1 - 2 = a T2

A

B

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A B C

20 km/h 25 km/h5 km/h

15–43.

The three freight cars A, B, and C have masses of 10 Mg, 5 Mg,and 20 Mg, respectively. They are traveling along the trackwith the velocities shown. Car A collides with car B first, fol-lowed by car C. If the three cars couple together after colli-sion, determine the common velocity of the cars after the twocollisions have taken place.

SOLUTIONFree-Body Diagram: The free-body diagram of the system of cars A and B when theycollide is shown in Fig. a. The pair of impulsive forces generated during the colli-sion cancel each other since they are internal to the system. The free-body diagramof the coupled system composed of cars A and B and car C when they collide isshown in Fig. b. Again, the internal pair of impulsive forces generated during thecollision cancel each other.

Conservation of Linear Momentum: When A collides with B, and then the coupledcars A and B collide with car C, the resultant impulsive force along the x axis is zero.Thus, the linear momentum of the system is conserved along the x axis. The initialspeed of the cars A, B, and C are

,

and

For the first case,

Using the result of and considering the second case,

Ans.vABC = -2.183 m>s = 2.18 m>s ;

(10000 + 5000)(4.167) + [-20000(6.944)] = (10000 + 5000 + 20000)vABC

(mA + mB)vAB + mC(vC)1 = (mA + mB + mC)vABCA :+ B

vAB

vAB = 4.167 m>s :

10000(5.556) + 5000(1.389) = (10000 + 5000)vAB

mA(vA)1 + mB(vB)1 = (mA + mB)v2A :+ B

AvC B1 = c25(103) mhd a

1 h3600 s

b = 6.944 m>s

AvB B1 = c5(103) mhd a

1 h3600 s

b = 1.389 m>s

AvA B1 = c20(103) mhd a

1 h3600 s

b = 5.556 m>s

F2

F1

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15–55.

SOLUTION

When the rope is stretched to its maximum, both the tug and barge have a commonvelocity. Hence,

Hence,

Ans.x = 0.221 m

454.282(103) + 0 = 439.661(103) +12

(600)(103)x2

T2 =12

(19 000 + 75 000)(3.059)2 = 439.661 kJ

T1 =12

(19 000)(4.167)2 +12

(75 000)(2.778)2 = 454.282 kJ

T1 + V1 = T2 + V2

v2 = 3.059 m>s19 000(4.167) + 75 000(2.778) = (19 000 + 75 000)v2

(:+ ) ©mv1 = ©mv2

(vB)1 = 10 km>h = 2.778 m>s(vT)1 = 15 km>h = 4.167 m>s

A tugboat T having a mass of 19 Mg is tied to a barge Bhaving a mass of 75 Mg. If the rope is “elastic” such that ithas a stiffness determine the maximumstretch in the rope during the initial towing.Originally both thetugboat and barge are moving in the same direction withspeeds and respectively.Neglect the resistance of the water.

1vB21 = 10 km>h,1vT21 = 15 km>h

k = 600 kN>m, (vB)1 (vT)1

B T

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