Lecture11

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Power

Average power Pavr=W/t Si unit: watt (W=J/s)

1hp=746 W



In electric power generation we use kWh

1kWh is energy transferred in 1 h at constant rate of 1 kW=1000 J/s

3 3 6

1 kwh=(10 W)(3600 s)=( 10 J/s)(3600 s)=3.6x10 J



Ex.5.12 Power Delivered by an Elevator motor

3 2 3

A 10 -kg elevator carries a max. load of 8x10 -kg. A constant frictional force of 4x10 -N retards its motion

upward. What minimum power , in kw and in hp , must the motor deliver to lift the fully loaded elevator at a

constant speed of 3 m/s?

Fnet=ma=T+f+Mg=0 (v=const)



On y: T-f-Mg=0

3 3 2 3

T=f+ Mg=4x10 N+(1.8x10 kg)(9.8 m/s )=2.16x10 N



Pavr=W/t=T* d *cos(0)/t=T*v (d/t=v)



Pavr = F*v valid for average and instantaneous power.

3 4

Pavr=(2.16x10 N)( 3 m/s)=6.48x10 W

B) What power must the motor deliver at the instant the speed of elevator is v if the motor is designed

2

to provide the elevator car with an upward acceleration of 1 m/s ?



Fnet=ma=T+f+Mg=Ma



On y: T-f-Mg=Ma



T=f+Mg+Ma



Pinst = T*v

Ex. 5.13 Shamu sprint

Whale has mass 8000kg, a=30 mi/h. Find whale Pavr it would need to generate to reach a speed of 12 m/s

in 6 s.(neglect water resistance).

Pavr=W/t



W=K work-kinetic theorem

2 2 2 2 6

K= mvf /2- mvi /2=1/2*8x10 x(12 m/s ) -0=5.76x10 J

6

Pavr=W/t = 5.76x10 J/6 s

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Ex.5.14 Speedboat power

3

What Pengine would a 1x10 –kg speedboat need to go from rest to 20 m/s in 5 s.? Water exerts constant drag

2

force fd=5x10 N (a=const) \



Pengine= W engine/t



W net =K

2 2

W net =K = mvf /2- mvi /2



2 2

W net = W engine +W drag= mvf /2- 0 W engine=W net -W drag= mvf /2- W drag

vf=vi+at=0+at

2

20 m/s=a(5 s)a=4 m/s



vf2- vi2 =2ax

2 2

(20 m/s) - 0=2(4 m/s )xx=50 m



2 4

W drag = - fdx= - (5x10 )(50 m)= -2.5x10 J

2 3 2 4

W engine=W net -W drag= mvf /2- W drag =1/2(10 kg)(20m/s) -( -2.5x10 J)

5

W engine=2.25x10 J

5 4

Pengine= W engine/t=2.25x10 J/5 s=4.5x10 W=60.3 hp





The Isolated System conservation of mechanical energy E mech =0 (isolated system with no

nonconservative force acting) (E mech =K+U)



Ex.8.1 Ball in Free Fall (Book2 ch.8-1,page 200)

A ball of mass m is dropped from a height h above the ground

a)Find the speed of the ball when it is at height y above the ground.

Ei = Ef



E i = K i +U i= K f +U f



mvi2/2+mgh= mvf2/2+mgy



0+mgh= mvf2/2+mgy

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vf2=2g(h-y) vf= (2g(h-y)) ½



b)Find the speed of the ball when it is at height y above the ground if

at the instant of release it has upward vi at the height h above the ground.



mvi2/2+mgh= mvf2/2+mgy



vf2= vi2+2g(h-y) ) vf= (vi2+2g(h-y)) ½



EX.8.2 A grand Entrance

Apparatus to support an actor m=65-kg to “fly” during the performance.

Sandbag M=130kg ( never lift above the floor), R=3m length of the cable to first pulley

angle from vertical line. What is the max value of can have before the sandbag lifts off the floor

W net = W T + W g =K

W net = 0 + W g =K  conservation of energy



mvi2/2+mgyi= mvf2/2+mgyf



0+mgyi= mvf2/2+0



yi=R-RcosR(1- cos



vf2Rg(1- cos vfRg(1- cos1/2



T+Ff Fcen



0n y: T-mg=m vf2R



T=mg+m vf2R

sandbag at rest T=Mg





Mg=mg+m vf2R



Mg-mg=m vf2R=mRg(1- cosR= mg(1- cos



cosmgmo





Ex. 8.3 The Spring –Loaded Popgun

A popgun consists of a spring of unknown spring constant. When the spring is compressed 0.12m, the

gun, when fired vertically, is able to launched a 35-g, projectile to a max. height of 20m above the

position of the projectile as it leaves the spring.

a) neglect all resistive forces, find the spring constant.

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Ug=Us=0 when the projectile leaves the spring yi= - 0.12m, yf= 20m





W net = W s+ W g =K  conservation of energy



Ei = Ef



E i = K i +U ig+U is= K f +U fg+U fs



mvi2/2+mgyi+1/2kyi2= mvf2/2+mgyf+1/2kyf2



0+mgyi+1/2kyi2= 0+mgyf+0



k=2mg(yf-yi)/y i2=2(0.035kg)(9.8m/s2)[20m-( - 0.12m)]/( -0.12m) 2=958 N/m



b) find the speed of the projectile as it moves through the equilibrium position of the

spring Ug=Us=0 y=0

E i = K i +U ig+U is= K f +U fg+U fs



mvi2/2+mgyi+1/2kyi2= mvf2/2+mgyf+1/2kyf2



0+mgyi+1/2kyi2= mvf2/2+0+0



vf2= 2gyi+1/(2m)kyi2



vf= (2gyi+1/(2m)kyi2) 1/2=(2(9.8m/s2 )( - 0.12) 2+(958 N/m)/(2(0.035kg)) 1/2=19.8m/s





Situation involving Kinetic friction

W net = W kfric +W s+ W g =K



W net = - f k d+W s+ W g =K



0=f k d-W s- W g +K= E int +Ug +Us +K=E intE



Ex.8.4 A block pulled on a rough surface

A 6-kg block initially at rest is pulled to the right along a horizontal surface by a constant horizontal

force of 12 N.

a) find the speed of the block after it has moved 3m if the surfaces in contact have a coefficient of

kinetic friction of 0.15

W net = - f k d+W app+ W N + W g =K



W net = - f k d+Fd+ 0+0=K

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2 2 2

W net = - f k d+Fd= mvf /2- mvi /2= mvf /2-0



2

W net = -  k mgd+Fd= mvf /2 mul. by 2/m



vf2=2Fd/m- k gd



vf=(2Fd/m- k gd) 1/2=[2(12N)(3m)/(6kg)-2(.15)(9.8m/s2)(3m)] 1/2=1.8m/s









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