Work, springs, kinetic energy, and power

•  CAPA homework due Friday at 10pm

•  Midterm long answers have been graded and will be uploaded before recitations tomorrow.

•  You should address grading concerns with Prof. Dan Dessau.

Web page: http://www.colorado.edu/physics/phys1110/phys1110_sp12/

Announcements:

Review of work and kinetic energy

The work-energy theorem is

Defined kinetic energy as

Defined work as

Clicker question 1 Set frequency to BA

At the bowling alley, the ball-feeder mechanism must exert a force to push the bowling balls up a 1.0-m long ramp. The ramp leads the balls to a chute 0.5 m above the base of the ramp.

Approximately how much force must be exerted on a 5.0-kg bowling ball?

A: 200 N B: 50 N C: 25 N D: 5.0 N E: impossible to determine

Force*distance = work by feeder. Gravity does the negative of that (net work is ZERO) We've learned that work of gravity is -m*g*h, which here is -m*g*0.5 m. So the feeder must do F*distance = +m*g*0.5 m. The distance is 1.0 meter now, so F = +m*g*0.5 = 50 N *.5 = 25 N.

More on springs

relaxed spring – no force applied

extended spring

compressed spring

The force from a spring is given by where k is the spring constant. It determines how much the spring resists because the larger k is, the larger force you get for a given x.

Spring work

x

F

s

ks We learned that the area under the force vs distance curve gives the work and for the area is a triangle so for a displacement s, the work is

Last class we used the example of the force from a spring to motivate how to find the work done by a varying force

Can use the area under the curve technique to also determine displacement from the velocity vs time graph or change in velocity from the acceleration vs time graph

Spring loaded gun 20 N of force is used to push the spring in a gun 2 cm into the cocked position. A 20 g red paint ball is inserted and the spring is released with the gun pointing up. How fast does the ball leave the gun?

2 cm 20 N When the spring releases it will do

work on the paint ball equal to where s = 2 cm.

But what is k? 20 N of force compresses the spring 2 cm so from we find

This is the spring constant for this spring

Spring loaded gun 20 N of force compresses a spring in a gun by 2 cm. What speed will a 20 g ball leave the gun at? 2 cm

20 N

This is negligible compared to the work by the spring

and so the launch speed is

So and the initial kinetic energy is 0 so using the work-energy theorem:

But gravity will also do work on the ball

Demo of Spring Launcher

Work of Spring on ball = ½ kx2 = ½ mv2

Height a ball will go up neglecting friction and air drag is given by ;

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v f2 = vi

2 − 2g(x f − xi)

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h = x f − xi

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vi2 = 2gh

Therefore, the height of the ball after it is released is proportional to the work of the spring on the ball. Let’s see ….

Advantage of Work-Energy A block of mass 2 kg is attached to a spring whose spring constant is k=8 N/m. The block slides on an incline with

If the block starts at rest with the spring unextended, what is its speed when it has slid a distance d=0.5 m down the incline?

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µk =18;θ = 37°

Work done by gravity is positive, by spring and friction is negative!

The work done by each of the forces on the block is

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Wg = m g • s = +mgd sinθ

W f = f • s = −µk (mgcosθ)d

Wsp = −12

kd2

The work-energy theorem says

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ΔK =12mv 2 − 0

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mgd sinθ −µk (mgcosθ)d −12kd2 =

12mv 2 Answer v=2 m/s

Clicker question 2 Set frequency to BA A hockey puck sliding on an ice rink at 1 m/s slides onto a rug someone left on the ice. The puck comes to rest after moving 1 m on the carpet. How far along the rug would the puck go, if its initial speed was 2 m/s? A.  1.5 m B.  2 m C.  3 m D.  4 m E.  Impossible to determine

Apply the Work-Energy Theorem. If the puck slides twice as far, the friction does twice as much (negative) work.

Doubling the speed quadruples the kinetic energy So the work by friction to reduce the kinetic energy to 0 (make it stop) must also increase by a factor of 4. Force is constant so the distance must increase by a factor of 4.

Another Problem using Work-Energy A crate of mass m is dropped onto a conveyer belt that moves at a constant speed v. The coefficient of kinetic friction is .

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µk

1) What is the work done by friction?

Since friction and displacement are in same direction work by friction is positive and

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Wf = µkmgd = ΔK =12mv 2

2) How far does the crate move before reaching its final speed?

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f = µkN = µkmg

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d = v 2 /(2µkg)

3) When the crate reaches its final speed, how far has the belt moved?

If the crate takes a time t to reach speed v, then v = 0+at, where a is the crate acceleration. Distance travelled is d = ½ at2 or d =1/2 vt, but belt moves vt, so belt moves twice are far as crate when it is accelerating.

Some notes on work and energy

Mechanical energy dissipated by friction means the same as the work done by friction. Remember that an amount of energy must be positive.

Energy is positive. For instance .

The change of energy can be either positive or negative

Power Work is a force applied over a distance.

But shouldn’t there be some extra credit for doing the work in a short amount of time?

Isn’t it strange that walking up 10 flights of stairs is the same amount of work as running up 10 flights of stairs?

New concept: Power is work divided by time.

Average power is

Instantaneous power is

Power continued Remember work is force times distance so for a constant force F acting in the same direction as the displacement Δx the average power is If we allow the force to be in any direction and let then the instantaneous power is

So we have two ways to determine power:

Instantaneous power:

Average power: which for a force along the velocity vector is

Power units Since power is work divided by time, the unit is joules/second (J/s) which is called a watt (W). So 1 W = 1 J/s.

Be careful not to confuse work with watt (both are W)

The other common power unit is horsepower (hp). 1 hp = 746 W

Originally developed to compare steam engines to draft horses

Example 1 (with Power)

A pump raises water from a well of depth 20m at a rate of 10 kg/s and discharges it at 6 m/s. What is the power of the motor?

The pump does work to lift the water and also to change its kinetic energy. The total work on the mass m would be

Since h + g are constant, the instantaneous power is

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W = mgh +12mv 2

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P =dWdt

=dmdt

gh +v 2

2

P = (10kg /s)[(9.8m /s2)(20m) +12(6m /s)2]

P ≈ 2180W

Example 2 (with Power) A 103-kg car requires 12 hp to cruise at a steady 80km/h on a level road. What would be the power required to move up a 10 degree incline at the same speed? Assume the total frictional force due to the road and air resistance is fixed.

At constant speed the wheels (actually the road) provide a forward force F that just balances the rolling friction fr of the tires and the drag force fd due to the air. This means that fT =fr + fd. Hence the power delivered to the wheels of the car is P = fTv. Converting to SI units gives v=80 km/h=(80 x 103 m)/(3600 s) = 22.2 m/s and 12 hp = 12 x 746 =8.95 x 103 W. Thus,

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f =Pv

=8.95 ×103W22.2m /s

= 403N

Now ready for the incline …………..

Example 2 cont.

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fT =Pv

=8.95 ×103W22.2m /s

= 403N

To go up an incline at constant speed, requires the car to have a force . The power required is

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F = fT + mgsin10°€

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P = F •v = (403N +103kg•9.8m /s2 • sin10°)(22.2m /s)P = 46.6 ×103W = 62.7hp

Note this power is 5x higher than on a level road!