**PHY 1151**

__Chapter 7; Work and Kinetic Energy__

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Chapter 7 | Course Calendar
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**D7.1 **How much work
is done by gravity on a skier weighing 600 N who slides 500 m down a hill
(measured along the hill) which makes an angle of 30° with the horizontal
as sketched in the figure?

**W = F D cos
= F D cos 60° = (600 N) (500 m) (0.50) = 150,000 J**

**
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D7.2

a) How much work is done on the puck by the stick as it moves a distance of 0.7 m?

b) What will be the puck's speed after it has moved that 0.7 m, assuming it started at rest?

c) From there the puck leaves the stick and slides across the floor. How far will it slide?

a)W = F D cos = (25 N) (0.7 m) (0.766) = 13.4 JRemember, this is the work done

by the stick. It isnotthe total or net work done on the puck (for the friction forcealsodoes work on the puck; however the weight and the normal force donotdo any work on the puck).b) Now we need the

totalwork done on the puck.W_{tot}= W_{net}= W_{stick }+ W_{f}+ W_{N}+ W_{w}We just found the work done by the stick,

W_{stick}= 13.4 J

W_{f}= F_{f}D cos 180° = F_{f}(0.7 m) (- 1)Now we must pause and determine the friction force F

_{f},F_{f}= µ F_{N}= (0.10) F_{N}And that means we must find the normal force, F

_{N},

The sum of all the y-components of the forces must be zero since there is no acceleration in the y-direction,F_{y}= - (25 N)(0.643) + F_{N}- 2.94 N = 0

F_{N}= 2.94 N + 16.07 N = 19.01 N

F_{f}= µ F_{N}= (0.10) F_{N}= (0.10) (19.01 N) = 1.9 N

W_{f}= (1.9 N) (0.7 m) ( - 1) = - 1.33 JThe work done by the weight and by the normal force are both

zerosince these forces areperpendicularto the distance moved,W_{N}= 0

W_{w}= 0Therefore,

W_{tot}= W_{net}= W_{stick}+ W_{f}+ W_{N}+ W_{w}

W_{tot}= W_{net}= 13.4 J - 1.33 J + 0 + 0

W_{tot}= W_{net}= 12.1 JWhat does this amount of work

cause? From the work-energy theorem (that is, from conservation of energy), we know that work done on an object causes a change in its kinetic energy; that is,W = KE_{f}- KE_{i}Since this puck starts at rest, we know

KE_{i}= 0so that

W = KE_{f}= (1/2) m v_{f}^{2}= (^{1}/_{2}) (0.3 kg) v_{f}^{2}= 12.1 J

v_{f}^{2}= 12.1 J/0.15 kg = 80.7 m^{2}/s^{2}

v_{f}= 9.0 m/s

c) Now what happens? At the end of the 0.7 m the stick no longer exerts a force and work is doneonly by the friction force. The puck starts out with 12.1 J of KE and finally comes to rest a distance D from where the stick released it. Again, we apply the work-energy theorem,W = - F_{f}D = - (1.9 N) D

W = KE_{f}- KE_{i}= 0 - 12.1 J

- ( 1.9 N ) D = - 12.1 J

D = ( 12.1 / 1.9 ) m

Cost = (1284 kW-h)($0.115/kW-h) = $147.66

E = 1284 kW-h = 1284 x 10^{3}W-h [^{(J/s)}/_{W}] (^{3600 s}/_{h })

E = 4.62 x 10^{9}J

E = P t = (0.25 mW) (1 h) () () () = 0.9 J

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P = F v (assuming F and v are parallel)

**P = F v**

As usual, we must change the speed from units of km/h to units of m/s

F = =

F = 1980 N

This is F_{water}, the force the water exerts on the
boat.

When a skier is towed the power is 15% greater,

**Now the net force is**

and that is the force that goes into

F = = = 2277 N

F

F

This is the tension in the ski tow rope

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As usual, we must change the speed from units of km/h to units of m/s

KE = (

KE = (

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**mass (kg)**
**speed (m/s)**
**A**
**0.225 kg**
**0.85 m/s**
**B**
**0.225**
**0.90**
**C**
**0.250 **
**0.35**
**D**
**0.250**
**0.50**
**E**
**0.267**
**0.65**
**F**
**0.315**
**0.82**
**G**
**0.450**
**0.45**
**H**
**0.450**
**0.52****KE = ( ^{1}/_{2})
m v^{2}**

A) KE = (^{1}/_{2}) (0.225) (0.85 ^{m}/_{s})^{2} = 0.0813 J

B) KE = (^{1}/_{2}) (0.225) (0.90 ^{m}/_{s})^{2} = 0.0911 J

C) KE = (^{1}/_{2}) (0.250) (0.35 ^{m}/_{s})^{2} = 0.0153 J

D) KE = (^{1}/_{2}) (0.250) (0.50 ^{m}/_{s})^{2} = 0.0313 J

E) KE = (^{1}/_{2}) (0.267) (0.65 ^{m}/_{s})^{2} = 0.0564 J

F) KE = (^{1}/_{2}) (0.315) (0.82 ^{m}/_{s})^{2} = 0.1059 J

G) KE = (^{1}/_{2}) (0.450) (0.45 ^{m}/_{s})^{2} = 0.0456 J

H) KE = (^{1}/_{2}) (0.450) (0.52 ^{m}/_{s})^{2} = 0.0608 J

**
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**v =
t
=
=
=
= 585 s
Energy = (0.32 kW)(585 s)
Energy = (320 W)(585 s)(
)
Energy = 187,000 J**

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Chapter 7| Course Calendar
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**(c) 2005, Doug Davis; all rights reserved.**