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14–1.

SOLUTIONEquation of Motion: Since the crate slides, the friction force developed between thecrate and its contact surface is . Applying Eq. 13–7, we have

Principle of Work and Energy: The horizontal component of force F which actsin the direction of displacement does positive work, whereas the friction force

does negative work since it acts in the opposite direction to that of displacement. The normal reaction N, the vertical component of force F and the weight of the crate do not displace hence do no work. Applying Eq.14–7,we have

Ans.v = 10.7 m s

-L

25 m

15 m36.55 ds =

12

(20)v2

12

(20)(8 2) +L

25 m

15 m100 cos 30° ds

T1 + aU1 - 2 = T2

Ff = 0.25(146.2) = 36.55 N

N = 146.2 N

+ caFy = may ; N + 100 sin 30° - 20(9.81) = 20(0)

Ff = mkN = 0.25N

30°

FThe 20-kg crate is subjected to a force having a constant direction and a magnitude F = 100 N. When s = 15 m, the crate is moving to the right with a speed of 8 m/s. Determine its speed when s = 25 m. The coefficient of kinetic friction between the crate and the ground is mk = 0.25.

Ans:v = 10.7 m>s

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14–3.

The crate, which has a mass of 100 kg, is subjected to theaction of the two forces. If it is originally at rest, determinethe distance it slides in order to attain a speed of Thecoefficient of kinetic friction between the crate and thesurface is .mk = 0.2

6 m>s.

SOLUTION

Equations of Motion: Since the crate slides, the friction force developed betweenthe crate and its contact surface is . Applying Eq. 13–7, we have

Principle of Work and Energy: The horizontal components of force 800 N and1000 N which act in the direction of displacement do positive work, whereas thefriction force does negative work since it acts in theopposite direction to that of displacement. The normal reaction N, the verticalcomponent of 800 N and 1000 N force and the weight of the crate do not displace,hence they do no work. Since the crate is originally at rest, . ApplyingEq. 14–7, we have

Ans.s = 1.35m

0 + 800 cos 30°(s) + 1000a45bs - 156.2s =

12

(100) A62 B

T1 + aU1-2 = T2

T1 = 0

Ff = 0.2(781) = 156.2 N

N = 781 N

+ c ©Fy = may; N + 1000a35b - 800 sin 30° - 100(9.81) = 100(0)

Ff = mk N = 0.2N

34

5

1000 N

30

800 N

Ans:s = 1.35 m

381

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14–5.

Determine the required height h of the roller coaster so that when it is essentially at rest at the crest of the hill A it will reach a speed of 100 km>h when it comes to the bottom B. Also, what should be the minimum radius of curvature r for the track at B so that the passengers do not experience a normal force greater than 4mg = (39.24m) N? Neglect the size of the car and passenger.

Solution

100 km>h =100(103)

3600= 27.778 m>s

T1 + ΣU1 - 2 = T2

0 + m(9.81)h =12

m(27.778)2

h = 39.3 m Ans.

+ cΣFn = man; 39.24 m - mg = ma(27.778)2

rb

r = 26.2 m Ans.

A

h

B

r

Ans:h = 39.3 mr = 26.2 m

387

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14–11.

The force of F = 50 N is applied to the cord when s = 2 m. If the 6-kg collar is orginally at rest, determine its velocity at s = 0. Neglect friction.

SolutionWork. Referring to the FBD of the collar, Fig. a, we notice that force F

does positive work but W and N do no work. Here, the displacement of F is

s = 222 + 1.52 - 1.5 = 1.00 m

UF = 50(1.00) = 50.0 J

Principle of Work And Energy. Applying Eq. 14–7,

T1 + ΣU1 - 2 = T2

0 + 50 =12

(6)v2

v = 4.082 m>s = 4.08 m>s Ans.

A

s

1.5 m

F

Ans:v = 4.08 m>s

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14–13.

SOLUTION

Ans.

Solving for the positive root,

Ans.

Ans

Ans.

.vC = 54.1 ft>s

12a 2

32.2b(5)2 + 2(45) =

12a 2

32.2bv2

C

TA + ©UA-C = TC

d = 31.48a45b(0.89916) = 22.6 ft

t = 0.89916 s

16.1t2 + 18.888t - 30 = 0

30 = 0 + 31.48a35b t +

12

(32.2)t2

A + T B s = s0 + v0t -12

ac t2

d = 0 + 31.48a45b t

a :+ b s = s0 + v0t

vB = 31.48 ft>s = 31.5 ft>s

12a 2

32.2b(5)2 + 2(15) =

12a 2

32.2bv2

B

TA + ©UA-B = TB

The 2-lb brick slides down a smooth roof, such that when itis at A it has a velocity of Determine the speed of thebrick just before it leaves the surface at B, the distance dfrom the wall to where it strikes the ground, and the speedat which it hits the ground.

5 ft>s.

30 ft

d

A

B

15 ft

5 ft/s

5

x

y

3

4

Ans:vB = 31.5 ft>sd = 22.6 ftvC = 54.1 ft>s

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14–17.

F � 30 lbA

CB

y y � x2

x

4.5 ft

3 ft1 ft 2 ft

12

If the cord is subjected to a constant force of lb andthe smooth 10-lb collar starts from rest at A, determine itsspeed when it passes point B. Neglect the size of pulley C.

F = 30

SOLUTION

Free-Body Diagram: The free-body diagram of the collar and cord system at anarbitrary position is shown in Fig. a.

Principle of Work and Energy: By referring to Fig. a, only N does no work since italways acts perpendicular to the motion. When the collar moves from position A toposition B, W displaces upward through a distance , while force F displaces a

distance of . The work of F is positive,whereas W does negative work.

Ans.vB = 27.8 ft>s

0 + 30(5.5) + [-10(4.5)] =12

a 1032.2bvB

2

TA + gUA - B = TB

s = AC - BC = 262 + 4.52 - 2 = 5.5 ft

h = 4.5 ft

Ans:vB = 27.8 ft>s

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14–23.

The 8-kg block is moving with an initial speed of 5 m>s. If the coefficient of kinetic friction between the block and plane is mk = 0.25, determine the compression in the spring when the block momentarily stops.

SolutionWork. Consider the force equilibrium along y axis by referring to the FBD of the block, Fig. a

+ cΣFy = 0; N - 8(9.81) = 0 N = 78.48 N

Thus, the friction is Ff = mkN = 0.25(78.48) = 19.62 N and Fsp = kx = 200 x. Here, the spring force Fsp and Ff both do negative work. The weight W and normal reaction N do no work.

UFsp= - L

x

0 200 x dx = -100 x2

UFf= -19.62(x + 2)

Principle of Work And Energy. It is required that the block stopped momentarily, T2 = 0. Applying Eq. 14–7

T1 + Σ U1 - 2 = T2

12

(8)(52) + ( -100x2) + [-19.62(x + 2)] = 0

100x2 + 19.62x - 60.76 = 0

Solved for positive root,

x = 0.6875 m = 0.688 m Ans.

Ans:x = 0.688 m

2 m 5 m/s

kA � 200 N/m

AB

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*14–28.

The 10-lb box falls off the conveyor belt at 5-ft>s. If the coefficient of kinetic friction along AB is mk = 0.2, determine the distance x when the box falls into the cart.

SolutionWork. Consider the force equilibrium along the y axis by referring to Fig. a,

+ cΣFy′ = 0; N - 10 a45b = 0 N = 8.00 lb

Thus, Ff = mkN = 0.2(8.00) = 1.60 lb. To reach B, W displaces vertically downward 15 ft and the box slides 25 ft down the inclined plane.

Uw = 10(15) = 150 ft # lb

UFf= -1.60(25) = -40 ft # lb

Principle of Work And Energy. Applying Eq. 14–7

TA + Σ UA - B = TB

12

a 1032.2

b (52) + 150 + (-40) =12

a 1032.2

b vB2

vB = 27.08 ft>s

Kinematics. Consider the vertical motion with reference to the x-y coordinate system,

( + c ) (SC)y = (SB)y + (vB)yt +12

ay t2;

5 = 30 - 27.08 a35b t +

12

(-32.2)t2

16.1t2 + 16.25t - 25 = 0

Solve for positive root,

t = 0.8398 s

Then, the horizontal motion gives

S+ (Sc)x = (SB)x + (vB)x t ;

x = 0 + 27.08 a45b(0.8398) = 18.19 ft = 18.2 ft Ans.

5 ft/s

30 ft

5 ft

15 ft

B

C

x

y

x

4

3 5A

Ans:x = 18.2 ft

420

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*14–44.

SOLUTIONEquation of Motion: The force F which is required to maintain the car’s constantspeed up the slope must be determined first.

Power: Here, the speed of the car is .

The power output can be obtained using Eq. 14–10.

Using Eq. 14–11, the required power input from the engine to provide the abovepower output is

Ans.=66.418

0.65= 102 kW

power input =power output

e

P = F # v = 2391.08(27.78) = 66.418(103) W = 66.418 kW

y = B100(103) m

hR * a 1 h

3600 sb = 27.78 m>s

F = 2391.08 N

+©Fx¿ = max¿; F - 2(103)(9.81) sin 7° = 2(103)(0)

An automobile having a mass of 2 Mg travels up a 7° slopeat a constant speed of If mechanical frictionand wind resistance are neglected, determine the powerdeveloped by the engine if the automobile has an efficiencyP = 0.65.

v = 100 km>h.7�

Ans:power input = 102 kW

424

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*14–48.

SOLUTIONPower: The work done by the man is

U = Wh = 150(15) = 2250 ft # lb

Thus, the power generated by the man is given by

Pman =Ut=

22504

= 562.5 ft # lb>s = 1.02 hp Ans.

The power of the bulb is Pbulb = 100 W * a 1 hp

746 Wb * a550 ft # lb>s

1 hpb

= 73.73 ft # lb>s. Thus,

t =UPbulb

=225073.73

= 30.5 s Ans.

The man having the weight of 150 lb is able to run up a 15-ft-high � ight of stairs in 4 s. Determine the power generated. How long would a 100-W light bulb have to burn to expend the same amount of energy? Conclusion: Please turn off the lights when they are not in use! 15 ft

Ans:Pman = 1.02 hpt = 30.5 s

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14–54.

SOLUTION

Ans.Pi =PO

e=

423.00.68

= 622 kW

PO = F # v = [13.8(103) + 10(30)](30) = 423.0 kW

v = 0 + 6(5) = 30 m>s

( :+ ) v = v0 + ac t

F = 13.8(103) + 10 v

:+ ©Fx = m ax; F - 10v = 2.3(103)(6)

The sports car has a mass of 2.3 Mg and accelerates atstarting from rest. If the drag resistance on the car

due to the wind is where v is the velocity inm/s, determine the power supplied to the engine when

The engine has a running efficiency of P = 0.68.t = 5 s.

FD = 110v2 N,6 m>s2,

FD

Ans:Pi = 622 kW