Power

Other Forms of Energy

Power – an example

Two cars drive up a hill. My 1991 Ford Escort takes on your 2013 Ford F150 pickup. They both make it to the top. How does the change in energy of the Escort compare to the change in energy of the F150 (assume they have equal masses)?

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ΔE = mgh f − mghi

= mgh

Both cars drive up the same vertical distance, so both cars have the same change in energy. But things are clearly not equal…

The pickup manages to change its energy in much less time than the Escort.

Power defined

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P ≡W

Δt=

FΔx

Δt= Fvavg

Power is the rate at which work is done (alternatively, rate at which energy is used).

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P[ ] =W

t

⎡ ⎣ ⎢

⎤ ⎦ ⎥=

kg m2

s3 = Watt

Some common items:Light bulb (incandescent): 40 – 100 WLight bulb (CFL): 9 – 30 WHair dryer: 1200 WFord F150: 224,000 W

This is NOT what you pay for from the electric company, despite the similar sounding name (kWhr).

Power – an example

What average power would a 1000 kg speedboat need to go from rest to 20 m/s in 5 seconds, assuming the water exerts a constant drag force of magnitude fd = 500 N and the acceleration is constant?

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P =W

Δt=

FbΔx

Δt

v f2 = v i

2 + 2aΔx

Δx =v f

2 −vi2

2a

a = ΔvΔt =

v f −vi

Δt

Fnet = ma = Fb − fd

Fb = ma + fd

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P =ma + fd( )

v f2 −vi

2

2a

⎛ ⎝ ⎜ ⎞

⎠ ⎟

Δt

=

m +fd Δt

v f − v i

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ v f

2 − v i2

( )

2Δt

=mv f

2 + fdv f Δt( )

2Δt

=mv f

2

2Δt+

fdv f

2=

KE f

Δt

⎛

⎝ ⎜

⎞

⎠ ⎟+ fdvavg( )

Momentum

Back to Newton…

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rF net = m

r a

r F net = m

Δr v

Δtr F netΔt = mΔ

r v

r F netΔt = m

r v f( ) − m

r v i( )

So we have a quantity, , that changes as the result of a force applied over a time interval.

This quantity is defined to be momentum.

If there is no net force on an object, then the object’s momentum is conserved.

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mr v

This equation is called the impulse equation. It tells us what to when an object is subjected to a net force.

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rp ≡ m

r v

Impulse ExampleA car approaches a tree at 25 m/sec (55mi/hr). What force is needed to stop

the 60 kg driver in 0.1 sec? Accident investigators, based upon crash tests, believe that an unrestrained body will ceasing moving forward within 0.1 sec of striking dashboard.

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rF Δt = mΔ

r v

F = mΔv

Δt

= 6025 − 0

.1

⎛

⎝ ⎜

⎞

⎠ ⎟=15,000 N

≈ 3,400 lbs

What about if they wear a seatbelt? Then they decelerate in ~0.5 sec.

Impulse – Area under F vs t graphImpulse (J) is the area of the force vs time graph, and also equal to mΔv

J = Area = 5(10) = 50Nm

If m = 2.5, then Δv = 20m/s Area = ½ (10)10 = 50 NmIf m = 2.5, then Δv = 20m/s

Back to Newton…

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rF net = m

r a

r F net = m

Δr v

Δtr F netΔt = mΔ

r v

r F netΔt = m

r v f( ) − m

r v i( )

So we have a quantity, , that changes as the result of a force applied over a time interval.

This quantity is defined to be momentum.

If there is no net force on an object, then the object’s momentum is conserved.

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mr v

This formalism can be extended to a system of interacting objects. In that case,

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rp ≡ m

r v

€

rp total, f =

r p total, i

A freight car of mass m traveling at a speed of 3 m/s collides with a stationary freight car of equal mass. The two cars lock together and move off at what speed?

Two Freight Cars

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rp total, i = m1

r v 1,i + m2

r v 2,i

= mr v 0

r p total, f = m + m( )

r v f = 2m

r v f

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rp total, i =

r p total, f

mr v 0 = 2m

r v f

r v f =

m

2mr v 0

=1

2

⎛

⎝ ⎜

⎞

⎠ ⎟3 m/s

=1.5 m/s (to the right)

Dead in Its Tracks

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rp leopard = mL

r v L = (40 kg)(10 m/s) = 400 Nsec

r p bullet = mB

r v B = (0.06 kg)(900 m/s) = 54 Nsec

40054 = 7.4 bullets (8 bullets)

A hunter has a rifle that shoots 60 gm bullets at a speed of 900 m/sec. A 40 kg leopard springs at him with a speed of 10 m/sec. How many bullets must the hunter fire in order to stop the leopard in its tracks?

2D Problem500 kg50 m/s

200kg30 m/s

???

25000Ns

~27000 Ns

6000Ns

27000Ns/700kg =38.6m/s