Ch 7, Work and Energy

Ch7; 2, 6, 11, 20, 21, 28, 40, 45, 52, 53Homework:## Questions 4, 5, 9, 12, 14

Additional problems from Serway's fourth edition

(4 ed) 7.1 A cheerleader lifts his 50.0-kg partner straight up off the ground a distance of 0.60 m before releasing her. If he does this 20 times, how much work has he done?

(4 ed) 7.2 If you push a 40.0-kg crate at a constant speed of 1.40 m/s across a horizontal floor (_{k}= 0.25), at what rate

(a) is work being done on the crate by you and

(b) is energy dissipated by the frictional force?

(4 ed) 7.3 A block of massmhangs on the end of a cord and is connected to a block of massMby the pulley arrangement shown in Figure P7.41. Using energy considerations,

(a) find an expression for the speed ofmas a function of the distance it has fallen. Assume that the blocks are initially at rest and there is no friction.

(b) Repeat (a) assuming sliding friction (_{k}) between massMand the table.

(c) Show that the result obtained in (b) reduces to that obtained in (a) in the limit as_{k}goes to zero.

(4 ed) 7.4 A block of mass 0.60 kg slides 6.0 m down a frictionless ramp inclined at 20^{o}to the horizontal. It then travels on a rough horizontal surface where_{k }= 0.50.

(a) What is the speed of the block at the end of the incline?

(b) What is its speed after traveling 1.00 m on the rough surface?

(c) What distance does it travel on this horizontal surface before stopping?

(4 ed) 7.5 A time-varying net force acting on a 4.0-kg particle causes the particle to have a displacement given by

x = 2.0 t - 3.0 t^{2}+ 1.0 t^{3},

where x is in meters and t is in seconds. Find the work done on the particle in the first 3.0 s of motion.

(4 ed) 7.6 A 1,500-kg car accelerates uniformly from rest to 10 m/s in 3.0 s.

Find (a) the work done on the car in this time,

and (b) the average power delivered by the engine in the first 3.0 s,

and (c) the instantaneous power delivered by the engine at t = 2.0 s.

(4 ed) 7.7 A car of weight 2,500 N operating at a rate of 130 kW develops a maximum speed of 31 m/s on a level, horizontal road. Assuming that the resistive force (due to friction and air resistance) remains constant.

(a) What is the car's maximum speed on an incline of 1 in 20 (ie,if theta is the angle of the incline with the horizontal, sin = 1/20)?

(b) What is its power output on a 1-in-10 incline if the car is traveling at 10 m/s?

Conceptual Questions

Q7.4 Can the kinetic energy of an object be negative?No. Kinetic energy is given by KE = (1/2) m v2. The mass m is always positive and v

^{2}is positive so KE must always be positive (of course, if v = 0 then KE = 0). KE cannotbe negative.

Q7.5 (a) If the speed of a particle is doubled, what happens to its kinetic energy?KE = (1/2) m v

^{2 so doubling the speed v means the Kinetic Energy is increased fourfold.}

(b) if the net work done on a particle is zero, what can be said about the speed?The net work done on a particle is equal to the

changein its Kinetic Energy. So if the net work iszerothe Kinetic Energy remainsconstantand that means thespeed, too, remainsconstant.

Q7.9 When a punter kicks a football, is he doing any work on the ball while his toe is in contact with it?Yes, a force is being exerted on the ball as it moves through a distance. That's the definition of work.

Is he doing any work on the ball after it loses contact with his toe?No, once his toe looses contact with the ball, the force his toe exerts is

zero.

Are any forces doing work on the ball while it is in flight?Yes,

gravityandair resistancecontinue to do work on the ball.

Q7.12 As a simple pendulum swings back and forth, the forces acting on the suspended mass are the force of gravity, the tension in the supporting cord, and air resistance.

(a) Which of these forces, if any, does no work on the pendulum?The force in the supporting cord is

perpendicularto the motion so that force doesno work.

(b) Which of thee forces does negative work at all times during its motion?Air resistance -- like all friction forces -- always does negative work.

(c) Describe the work done by the force of gravity while the pendulum is swinging.The force of gravity does

positive workas the pendulum "falls" from either extreme to its lowest position; this causes the pendulum tospeed up. The force of gravity doesnegative workas the pendulum "rises" from its lowest position to either extreme; this causes the pendulum toslow down.

Q7.14 An older model car accelerates from 0 to a speed v in 10 s. A newer, more powerful sports car accelerates from 0 to 2 v in the same time period. What is the ratio of powers expended by the two cars?Assume the two cars have the same mass.

The car going 2 v has

fourtimes the Kinetic Energy. If it acquired this much energy in the same time, then the power supplied to it is alsofourtimes as much.

Problems from the current (5th) edition of Serway and Beichner

7.2 A shopper in a supermarket pushes a cart with a force of 35 N directed at an angle of 25^{o}downward from the horizontal. Find the work done by the shopper as she moves down an aisle 50 m in length.W = F d cos

W = (35 N)(50 m) cos 25

^{o}W = (35 N)(50 m)(0.906)

W = 1 586 N-m = 1 586 J

It is entirely equivalent to think of this as

W = F

_{x}dW = [(35 m) cos 25

^{o}] (50 m)W = 1 586 J

7.6 A 15.0-kg block is dragges over a rough, horizontal surface by a 70.0-N force acting at 20^{o}above the horizontal. The block is displaced 5.0 m and the coefficient of kinetic friction is 0.30. Find the work done by

(a) the 70-N force,

(b) the normal force, and

(c) the force of gravity.

(d) What is the energy loss due to friction?W = F d cos

W

_{70}= F_{ext}d cos 20^{o}W

_{70}= (70 N)(5 m)(0.9397)

W_{70}= 328.9 JW

_{n}= n d cos 90^{o}

W_{n}= 0W

_{g}= w d cos 90^{o}

W_{g}= 0W

_{fr}= F_{fr}d cos 180^{o}What

isthe force of friction F_{fr? We do know that}F

_{fr}= 0.30 nBut what is the normal force

n? We see enough special cases when n = w = mg that one can intuitively jump to the wrong conclusion and thing that n = w = mg all the time. That isnotthe case in general and that isnotthe case here. The block is not jumping up off the plane and it is not burrowing down into the plane. Therefore, the sum of the forces in the vertical direction must be zero.F

_{net,y}= n + F_{ext,y}- w = 0n + (70 N)(sin 20

^{o}) - (15 kg)(9.8 m/s^{2}) = 0n + (70 N)(0.342) - (15)(9.8) N = 0

n + 23.9 N - 147 N = 0

n - 123.1 N = 0

n = 123.1 N

With the value of the normal force n now known, we can calculate the friction force F

_{fr}.F

_{fr}= 0.30 nF

_{fr}= 0.30 (123.1 N)F

_{fr}= 36.9 NNow we can calculate the work done by the friction force,

W

_{fr}= F_{fr}d cos 180^{o}W

_{fr}= - (36.9 N)(5.0 m)

W_{fr}= - 184.6 JWhat does the negative sign mean. The force of friction

decreasesthe energy. This is the energylosswhich shows up asheat.

7.11A forceF= (6i- 2j) N acts on a particle that undergoes a displacementd= (3i+j) m.Find (a) the work done by the force on the particle

and (b) the angle between

Fandd.W =

FdW = [(6

i- 2j) N] [(3i+j) m]W = [(6

i- 2j) (3i+j)] (N-m)W = [(6

i- 2j) (3i+j)] JThere are (at least)

twoways we can evaluate this dot-product.W = [(6

i) (3i) + (6i) (j) - (2j) (3i) - (2j) (j)] JW = [(6)(3)(

ii) + (6)(1)(ij) - (2)(3)(ji) - (2)(1)(jj)] JW = [(6)(3)(1) + (6)(1)(0) - (2)(3)(0) - (2)(1)(1)] J

W = [18 + 0 - 0 - 2] J

W = 16 JAn alternative is to remember the dot product in terms of components as

AB= A_{x}B_{x}+ A_{y}B_{y}Then the work can be written as

W =

Fd= F_{x}d_{x}+ F_{y}d_{y}W =

Fd= [(6)(3) + ( - 2)(1)] N-mW =

Fd= [18 - 2] J

W = 16 J

7.20 An archer pulls her bow string back 0.400 m by exerting a force that increases uniformly from zero to 230 N.

(a) What is the equivalent spring constant of the bow?

(b) How much work is done in pulling the bow?F = k x230 N = k (0.4 m)

k = 230 N / 0.4 m

k = 575 N/m

W = (

^{1}/_{2}) k x^{2}W = (

^{1}/_{2}) (575 N/m) (0.4 m)^{2}W = 46 J

7.21 A 6,000-kg freight car rolls along rails with negligible friction. The car is brought to rest by a combination of two coiled springs, as illustrated in Figure P7.21. Both springs obey Hooke's law, with k_{1}= 1600 N/m and k_{2}= 3400 N/m. After the first spring compresses a distance of 30.0 cm, the second spring (acting with the first) increases the force so that there is additional compression as shown in the graph. If the car is brought to rest 50.0 cm after first contacting the two-spring system, find the car's initial speed. The wording of this questionmayambiguous. When it asks "beIf the car is brought to rest 50.0 cm after first contacting the two-spring system, find the car's initial speed." -- exactly what is meant by "after first contacting the two-spring system"? This solution takes "after first contacting the two-spring system" to mean the car moves 50.0 cm after first contacting thespring, with kfirst_{1}= 1600 N/m.The first spring is compressed the full distance of x

_{1}= 0.50 m so it does work ofW

_{1}= (^{1}/_{2}) k_{1}x_{1}^{2}= 0.5 (1 600 N/m) (0.5 m)^{2}= 200 JThe second spring is compressed only after the freight car moves a distance of 0.30 m, so it is only compressed a distance of x

_{2}= 0.20 m so it does work ofW

_{2}= (^{1}/_{2}) k_{2}x_{2}^{2}= 0.5 (3 400 N/m) (0.2 m)^{2}= 68 JSo both springs have done a total amount of work of

W

_{Tot}= W_{1}+ W_{2}= 200 J + 68 J = 268 J[ There is an alternate way to do this. We could figure the work done during the first 0.30 m with k = 1600 N/m and then find the work done during the last 0.20 m from the "area" under the graph. The total amount of work should be the same, either way ].

Actually, this total work is

negativethat amount, W_{Tot}= - 268 J, since the force and displacement are in opposite direction. This has changed the freight car's Kinetic Energy from its initial value ofKE

_{i}= (^{1}/_{2}) m v_{i}^{2}to its final value of zero. That is

KE

_{i}= (^{1}/_{2}) m v_{i}^{2}= (^{1}/_{2}) (6 000 kg) v_{i}^{2}= 268 J = W_{springs}v

_{i}^{2}= 0.089 m^{2}/ s^{2}v

_{i}= 0.30 m/s

7.28 A 0.30-kg ball has a speed of 15.0 m/s.

(a) What is its kinetic energy?

(b) If its speed were doubled, what would be its kinetic energy?KE = (1/2) m v

^{2}KE = (1/2) (0.30 kg) (15.0 m/s)

^{2}

KE = 33.75 JSince the speed is

squared, doubling the speed will increase the KE by a factor offour.KE

_{dbl}= 4 KE_{orig}

KE_{dbl}= 135 J

7.40 An Atwood's machine has a 3.00 kg mass and a 2.00-kg mass at ends of the string (as in Figure 5.15). The 2.00-kg mass is released from rest on the floor, 4.00 m below the 3.00-kg mass.

(a) If the pulley is frictionless, what will be the speed of the masses when they pass each other?

(b) Suppose that the pulley does not rotate and the string must slide over it. If the total frictional force between the pulley and the string is 4.00 N, what are their speeds when the masses pass each other?

W_{net}= KEW

_{net}= W_{2}+ W_{3}= - (2 kg) (9.8 m/s^{2}) (2 m) + (3 kg) (9.8 m/s^{2}) (2 m)W

_{net}= 19.6 JThe w's in the diagram -- w

_{2}and w_{3}-- are the weights;w

_{2}= m_{2}g = (2.0 kg) (9.8 m/s^{2})w

_{3}= m_{3}g = (2.0 kg) (9.8 m/s^{2})The capital W's in the equation above -- W

_{2}and W_{3}-- are the amounts of work done on the 2.0-kg mass and the 3.0-kg massby gravity.Only the force of gravity does work on "the system". We need not -- can not -- consider the tension in the cord for that is an "internal force" just like the forces that hold the blocks together.

The blocks have a common velocity v,

KE = 0.5 (2.0 kg) v

^{2}+ 0.5 (3.0 kg) v^{2}= (2.5 kg) v^{2}KE = (2.5 kg) v

^{2}= 19.6 J = W_{net}v

^{2}= 7.84 m^{2}/ s^{2}v = 2.8 m / s

Now, consider

friction.W

_{f}= - (5.0 N) (2.0 m) = - 10 JW

_{net}= 19.6 J - 10 J = 9.6 JKE = (2.5 kg) v

^{2}= 9.6 J = W_{net}v

^{2}= 3.84 m^{2}/ s^{2}v = 1.96 m /s

7.45 A certain automobile engine delivers 2.24 x 10^{4}W (or 22.4 kW or 30 hp) to its wheels when moving at a constant speed of 27.0 m/s (about 60 mi/h or about 100 km/h). What is the resistive force acting on the automobile at that speed?P = F v

F = P / v

F = [2.24 x 10

^{4}W]/[27 m/s]

F = 830 N

7.52 An electron moves with a speed of 0.995 c.

(a) What is its kinetic energy?

(b) If you use the classical expression to calculate its kinetic energy, what percentage error would result?It is fun to look at more thanjustv = 0.995 c. Repetitive calculations are often easiest to do with a simple spreadsheet.

7.53 A proton in a high-energy accelerator moves with a speed of c/2. Using the work-energy theorem, find the work required toincreaseits speed to

(a) 0.750 c and

(b) 0.995 c.For common, ordinary speeds, we found that the (kinetic) energy of a particle with mass m and speed v is KE = (1/2) m v

^{2}. But forveryhigh speeds -- for speeds close to that of light -- we had to modify that and useE = m c

^{2}m = m

_{o}/ SQRT(1 - v^{2}/c^{2})E

_{o}= m_{o}c^{2 }KE = E - E

_{o }KE = m

_{o}c^{2}[ {1/SQRT(1 - v^{2}/c^{2})} - 1]This does

notlook very much like KE = (1/2) m v^{2}! So it is worth reminding ourselves tha this unusual looking expression does, indeed, reduce to KE = (1/2) m v^{2}for v << c !First, let's find the "rest energy" of the proton since this factor appears in all our future calculations.

E

_{o}= m_{o}c^{2 }E

_{o}= (1.672 x 10^{ - 27}kg)(3 x 10^{8}m/s)^{2 }E

_{o}= 5.016 x 10^{ - 11}JThis is perfectly okay as it is. However, instead of dealing with such small numbers, it is customary to talk about these energies in units of MeV -- millions of electron-Volts. We will explore MeV's more in PHY 1360 (next semester). For the present, we can simply pull out of the air the appropriate conversion factor,

1 eV = 1.6 x 10^{ - 19}J1 MeV = 1.6 x 10

^{ - 16}JTherefore,

E

_{o}= 5.016 x 10^{ - 11}J [1 MeV / 1.6 x 10^{ - 16}J]E

_{o}= 313 500 MevE

_{o}= 313.5 GevOne GeV is one giggaelectron-volt. Gigga means "billion" or 10

^{9}.E = m c

^{2}m = m

_{o}/ SQRT(1 - v^{2}/c^{2})E

_{orig}= [m_{o}/ SQRT(1 - v_{orig}^{2}/c^{2})] c^{2}E

_{orig}= [m_{o}c^{2}][1/ SQRT(1 - v_{orig}^{2}/c^{2})]E

_{orig}= [313.5 Gev][1/ SQRT(1 - {0.50 c}^{2}/c^{2})]E

_{orig}= [313.5 Gev][1/ SQRT(1 - {0.50}^{2})]E

_{orig}= [313.5 Gev][1/ SQRT(1 - 0.25^{})]E

_{orig}= [313.5 Gev][1/ SQRT(0.75^{})]E

_{orig}= [313.5 Gev][1/ 0.866]E

_{orig}= [313.5 Gev][1.155]E

_{orig}= 362 GevThat's the total energy -- KE plus "rest mass energy" -- that the proton has originally. Now, how much energy does it have for the speeds given in question (a) and question (b)?

E

_{a}= [m_{o}/ SQRT(1 - va^{2}/c^{2})] c^{2}E

_{a}= [m_{o}c^{2}][1/ SQRT(1 - v_{a}^{2}/c^{2})]E

_{a}= [313.5 Gev][1/ SQRT(1 - {0.750 c}^{2}/c^{2})]E

_{a}= [313.5 Gev][1/ SQRT(1 - {0.750}^{2})]E

_{a}= [313.5 Gev][1/ SQRT(1 - 0.5625^{})]E

_{a}= [313.5 Gev][1/ SQRT(0.4375^{})]E

_{a}= 313.5 Gev][1/ 0.661]E

_{a}= [313.5 Gev][1.512}E

_{a}= 474 GeVHow much energy do we need to

addto go from the original speed (v_{orig}= 0.500 c) tho this speed (v_{a}= 0.75 c)?E

_{a}= E_{a}- E_{orig}E

_{a}= (474 - 362) GeV

E_{a}= 112 GeVNow we do exactly the same thing for v

_{b}= 0.995 c.E

_{b}= [m_{o}/ SQRT(1 - v_{b}^{2}/c^{2})] c^{2}E

_{b}= [m_{o}c^{2}][1/ SQRT(1 - v_{b}^{2}/c^{2})]E

_{b}= [313.5 Gev][1/ SQRT(1 - {0.995 c}^{2}/c^{2})]E

_{b}= [313.5 Gev][1/ SQRT(1 - {0.995}^{2})]E

_{b}= [313.5 Gev][1/ SQRT(1 - 0.990025^{})]E

_{b}= [313.5 Gev][1/ SQRT(0.009975^{})]E

_{b}= [313.5 Gev][1/ 0.09987]E

_{b}= [313.5 Gev][10.012]E

_{b}= 3 139 GeVHow much energy do we need to

addto go from the original speed (v_{orig}= 0.500 c) tho this speed (v_{b}= 0.995 c)?E

E_{b}= E_{b}- E_{orig}_{b}= (3 139 - 362) GeV

E_{b}= 2 777 GeV

Solutions to the additional problems from Serway's fourth edition.

(4 ed) 7.1 A cheerleader lifts his 50.0-kg partner straight up off the ground a distance of 0.60 m before releasing her. If he does this 20 times, how much work has he done?w = mg = (50 kg) (9.8 m/s^{2}) = 490 NW = F s = (490 N) (0.60 m) = 294 N-m = 294 J

W

_{Tot}= 20 x W = (20) (290 J) = 5,880 JW

_{Tot}= 5,880 J

(4 ed) 7.2 If you push a 40.0-kg crate at a constant speed of 1.40 m/s across a horizontal floor (_{k}= 0.25), at what rate

(a) is work being done on the crate by you and

(b) is energy dissipated by the frictional force?P = F vWe must find the horizontal force F

From the diagram, we find that

F = F

_{f}F

_{n}= m g = (40 kg) (9.8 m/s^{2}) = 392 NF

_{f}= F_{n}= (0.25) (392 N) = 98 NF = 98 N

P = F v = (98 N) (1.4 m/s) = 137 N-m / s = 137 J / s = 137 W

The friction force F

_{f}dissipates energy at thesame rate,P = 137 W

(4 ed) 7.3 A block of massmhangs on the end of a cord and is connected to a block of massMby the pulley arrangement shown in Figure P7.41. Using energy considerations,

(a) find an expression for the speed ofmas a function of the distance it has fallen. Assume that the blocks are initially at rest and there is no friction.

(b) Repeat (a) assuming sliding friction (_{k}) between massMand the table.

(c) Show that the result obtained in (b) reduces to that obtained in (a) in the limit as_{k}goes to zero. When hanging massmmoves a distance x, the sliding massMmoves only half that distance, x/2. When hanging massmhas speed v, sliding massMhas a speed of only half that, v/2.When the hanging mass has moved a distance x from its initial position, the force of gravity has done work of

W

_{g}= m g xThat is the only work that is done on the "system" since the forces on sliding mass

M-- the force of gravity and the normal force of the table -- are perpendicular to the displacement. The tension in the cord is an "internal force" and does not act on the "system" from the outside. This work due to gravity is thenet workso it must equal the change in KE of the "system"W

_{g}= W_{net}= m g x = KE = KE - KE_{i}= KE - 0 = KE = KE_{M}+ KE_{m}m g x = (

^{1}/_{2}) M (^{1}/_{2 }v )^{2}+ (^{1}/_{2}) m v^{2}m g x =

^{1}/_{8 }M v^{2}+ (^{1}/_{2}) m v^{2}m g x = (

^{1}/_{2}) [^{1}/_{4 }M + m ] v^{2}v

^{2}= 2 m g x / [^{1}/_{4 }M + m ]v

^{2}= { 2 m g / [^{1}/_{4 }M + m ] } xThis is with

outfriction.

Now, go back andaddinfriction. With friction present, the friction force isF

_{f}= F_{N}= M gAs hanging mass

mmoves distance x, the sliding massMmoves a distance x/2 and work due to the friction force is done on it in the amount ofW

_{f}= - F_{f}(x/2) = - M g x /2That means the

network isW

_{net}= W_{g}+ W_{f}= m g x - M g x /2W

_{net}= m g x - M g x /2 = KE = KE - KE_{i}= KE - 0 = KE = KE_{M}+ KE_{m}m g x - M g x /2 = KE

_{M}+ KE_{m}m g x - M g x /2 = KE

_{M}+ KE_{m}= (^{1}/_{2}) M (^{1}/_{2 }v )^{2}+ (^{1}/_{2}) m v^{2}[ m - M / 2 ] g x = (

^{1}/_{2}) [ (^{1}/_{4}) M + m ] v^{2}v

^{2 }= 2 { [ m - M / 2 ] / [ (^{1}/_{4}) M + m ] } g xDoes this expression reduce to our earlier, friction

lessexpression (as itmust) when the coefficient of friction is zero? Set = 0 and see;v

^{2 }= 2 { [ m - M / 2 ] / [ (^{1}/_{4}) M + m ] } g xv

^{2 }= 2 { [ m - (0) M / 2 ] / [ (^{1}/_{4}) M + m ] } g xv

^{2 }= 2 { [ m ] / [ (^{1}/_{4}) M + m ] } g xv

^{2 }= 2 { [ m ] / [ (^{1}/_{4}) M + m ] } g xAnd, indeed, this is the same expression as we got for the friction

lesscase directly.

(4 ed) 7.4 A block of mass 0.60 kg slides 6.0 m down a frictionless ramp inclined at 20^{o}to the horizontal. It then travels on a rough horizontal surface where_{k }= 0.50.

(a) What is the speed of the block at the end of the incline?

(b) What is its speed after traveling 1.00 m on the rough surface?

(c) What distance does it travel on this horizontal surface before stopping?

(a) What is the speed of the block at the end of the incline?Onthe frictionlessincline, the normal force doesno work, W_{n}= 0. Only the force of gravity does any work,W

_{g}= F s cos = ( m g ) (6.0 m) cos 70^{o}= (0.60 kg) (9.8 m/s^{2}) (6.0 m) (0.342) = 12.1 JW

_{g}= 12.1 JNotice that this has the cosine of 70

^{o}rather than of 20^{o}. In writing W = F s cos , the angle is the anglebetweenthe force and the displacement.W

_{net}= W_{g}+ W_{n}= 12.1 JW

_{net}= KE = KE - KE_{i}= KEW

_{net}= (^{1}/_{2}) m v^{2}(

^{1}/_{2}) m v^{2}= (^{1}/_{2}) (0.60 kg) v^{2}= 12.1 Jv

^{2}= 40.22 m^{2}/ s^{2}v = 6.34 m / s

This is the speed at the bottom of the incline and the beginning of the rough horizontal plane that has friction.

From the force diagram, you can see that

F

_{n}= m g = (0.60 kg) (9.8 m/s^{2}) = 5.88 NF

_{f}= F_{n}= (0.5) (5.88 N) = 2.94 N, pointing to theleftAlong this rough, horizontal plane,

onlythis friction force F_{f}does any work. By the time the block has moved a distance of 1.0 m along the plane, the net work done on the block isW

_{net}= W_{f}= - (2.94 N) (1 m) = - 2.94 JW

_{net}= KE = KE_{1.0 m}- KE_{o}= KE_{1.0 m}- 12.1 J = - 2.9 JKE

_{1.0 m}= 12.1 J - 2.9 J = 9.2 JKE

_{1.0 m}= 9.2 J = (^{1}/_{2}) m v^{2}= (^{1}/_{2}) (0.60 kg) v^{2}v

^{2}= 30.7 m^{2}/ s^{2}v = 5.5 m / s

Now, how far does it travel before it completely stops. At that point, KE

_{f}= 0. The work done by friction in moving this distance X isW

_{net}= W_{f}= - (2.94 N) ( X_{f})And this is just equal to the change in KE, from its initial value of 12.1 J at the bottom of the inclined plane until it goes to zero,

- (2.94 N) X

_{f}= - 12.1 JX

_{f}= 4.11 m

(4 ed) 7.5 A time-varying net force acting on a 4.0-kg particle causes the particle to have a displacement given by

x = 2.0 t - 3.0 t^{2}+ 1.0 t^{3},

where x is in meters and t is in seconds. Find the work done on the particle in the first 3.0 s of motion.We need the change in KE so we need KE at t = 0 and KE at t = 3.0 sv = dx/dt = 2.0 t

^{0}- 3.0 ( 2 t ) + 1.0 (3 t^{2}) = 2.0 - 6.0 t + 3.0 t^{2}v

_{o}= v(t = 0) = 2.0 m/sv

_{f}= v(t = 3 s) = 2 - 6(3) + 3(3^{2}) = 2 - 18 + 27 = 11 m/sKE

_{o}= (^{1}/_{2}) m v_{o}^{2}= (^{1}/_{2}) (4.0 kg) (2 m/s)^{2}= 8 JKE

_{f}= (^{1}/_{2}) m v_{v}^{2}= (^{1}/_{2}) (4.0 kg) (11 m/s)^{2}= 242 JW

_{net}=KE = 234 J

(4 ed) 7.6 A 1,500-kg car accelerates uniformly from rest to 10 m/s in 3.0 s.

Find (a) the work done on the car in this time,

and (b) the average power delivered by the engine in the first 3.0 s,

and (c) the instantaneous power delivered by the engine at t = 2.0 s.W

_{net}=KE = 0.5 (1 500 kg) (10 m/s)^{2}= 75 000 J = 75 kJP

_{avg}= W / t = 75 kJ / 3 s = 25 kW = 25 000 WFor t = 2 s,

P = F v

F = m a = m (v / t) = [1 500 kg] [(10 m/s) / 3 s] = 5 000 N = 5 kN

a = v / t = (10 m/s) / 3 s = 3.33 m / s

^{2}v = v

_{i}+ a t = 0 + ( 3.33 m / s^{2}) (2 s) = 6.67 m / sP = (5 000 N) (6.67 m / s)

P = 33 350 W = 33.35 kW

(4 ed) 7.7 A car of weight 2,500 N operating at a rate of 130 kW develops a maximum speed of 31 m/s on a level, horizontal road. Assuming that the resistive force (due to friction and air resistance) remains constant.

(a) What is the car's maximum speed on an incline of 1 in 20 (ie,if theta is the angle of the incline with the horizontal, sin = 1/20)?

(b) What is its power output on a 1-in-10 incline if the car is traveling at 10 m/s?We know the power can be calculated from

P = F v

Since the car is not accelerating, the force F supplied by the engine is just equal to the resistive forces that retard its forward motion.

F = P / v = (130 000 W) / (31 m/s) = 4 194 N = 4.194 kN

This is the resistive force which we will

assumeto stay at this value.Now, climb a hill with a 1-in-20 incline.

On this hill,

in addition to the resistive forceof 4 194 N, there isalsoa component of the weight that points "down" the hill; let's call this force F_{||},F

_{||}= w sin = (2 500 N) (1 / 20) = 125 NNow the total retarding force is

F

_{Tot}= 4 194 N + 125 N = 4 319 NTo keep the car moving at constant speed, the engine must supply this

forceto the tires.P = F v

v = P / F = 130 000 W / 4 319 N = 30.09 m / s

v = 30.1 m / s

On a 1-in-10 incline, the component of the weight that is parallel to the road is

F

_{||}= w sin = (2 500 N) (1 / 10) = 250 NF

_{Tot}= 4 194 N + 250 N = 4 444 NP = F v = (4 444 N) (10 m / s) = 44 440 W = 44.44 kW

P = 44 440 W = 44.44 kW

| Hmwk, Ch 6 |

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