Work, Power and Energy

Work Work is done when a force (F) causes an object to move distance (d) in the direction of the force. One system transfers energy to another system equal to the amount of work done. Work (energy) is a scalar quantity. Work = Force x displacement W=Fd Units for work = Nm or Joules (J)

Work Quiz

Careful! the displacement must be in the same direction as the force (or a component of the force) applied. If the force is vertical and the displacement is horizontal, no work is done. If there is no F or d, no work is done. If part of the F is in the direction of the d, work is done.

Examples How much work is done when a horizontal 30 N force pushes a cart a distance of 3.5 meters? W=Fd W = (30 N)(3.5 m) W = 105 Nm

How much work is done lifting a 500 N barbell 1.5 m above the floor? W=Fd W=(500 N)(1.5m) W = 750 Nm

How much work is required to lift a 75 kg crate to a height of 2 m above the floor? W=Fd W= (75kg)(9.8m/s2)(2m) W=1470 kgm2/s2

A child pulls a wagon a horizontal distance of 5 m with a force of 200 N at an angle of 30o to the horizontal. How much work is done? W=Fd W=(200 N)( cos 30o)(5 m) W= 866 J

A 5 kg block sits at rest on a horizontal surface A 5 kg block sits at rest on a horizontal surface. A constant force of 3.0 N is applied to the block at an angle of 60o to the horizontal. How much work is done in sliding the block 6.0 m along the surface? W=Fd W=(3 N)(cos 60o)(6m) W= 9 Nm

For those of you just joining us Doing work on something is transferring energy to it. Work done = energy gained Work = Force x displacement The displacement must be multiplied by the force that made it move in that direction.

To lift something W=Fd W = (weight) (displacement) W= (mg) height W= (5kg)(9.8 m/s2)(3m) F= D=3m 5 kg

To push something along a horizontal surface W=Fd=(20N)(4 m) = 80 Nm Fapp=20 N D=4 m

To push or pull something using a force at an angle W = Fd W= (F cos θ )(d)

Power Power is the rate at which work is done. P= Work time Units are Nm/s or J/s or Watts (W) If Power = Work then . . . . . .

Power = Fd (work) t Power = Fv If the time it takes to do a given amount of work increases, the power decreases.

Examples A woman of mass 50 kg does 7.62 x 103 joules of work when she runs up three flights of stairs in 18.2 seconds to a point vertically above her starting point. What power does she develop? P=W/t = 7.62 x 103 J/18.2 sec=419 W What is her vertical distance? W = Fd 7.62 x 103 J = 490 N (d) d = 15.5 m

A constant horizontal force of 10 N is applied to a 2 kg block on a counter to overcome friction. How much power is dissipated when the box is moved 5.0 cm in 8.0 seconds? P=Fd/t = (10N)(.05 m) / 8s P=0.0625 W

A 4. 0 kg block sits at rest on a horizontal surface A 4.0 kg block sits at rest on a horizontal surface. A constant 10 N force is applied to the block at an angle of 40o to the horizontal. What power is utilized if the block moves a distance of 6.0 m in 5 seconds? P= Fd/t = (10N)(cos40)(6 m)/5 s=9.2 W

Review Work = Force x distance Work to lift = Force to lift x height lifted up = mg ∆ h Work done on an object = energy gained

Gravitational Potential Energy Potential energy = stored energy An object gains gravitational potential energy (∆ PEg) when work is done to lift it. ∆ PEg = mghf - mghi ∆ PEg = mg ∆ h Potential energy has the ability to do work. Gravitational potential energy is the energy an object has due to an elevated position. It loses this energy when it falls.

The work done to lift an object = the PEg gained by an object (∆ PEg) The work done to lift an object = the PEg gained by an object (∆ PEg). It is also the work the object can now do. So ∆ PEg= F d **(up) Which is ∆ PEg = mg ∆h

Objects with PEg have the ability to do work.

Examples How much gravitational potential energy is gained (i.e. how much work is done) when a 100 N boulder is lifted a vertical distance of 2.0 meters? W=Fd or PEg = mg ∆h W = (100 N)(2m) W = 200 J

How much gravitational potential energy does a 100 kg pile driver gain when it is lifted a vertical distance of 8 meters? W = Fd or PEg = mg ∆h W = (100 kg)(9.8 m/s2)(8m) W= 7840 J (It can now do 7840 J of work)

How much work is done when a 2 How much work is done when a 2.5 kg dumbbell is lifted from a height of 1.0 meter to 2.5 meters off the ground? W = Fd or ∆ PEg = mg ∆h ∆ PEg= (2.5 kg)(9.8 m/s2)(1.5 m) = 36.8 J

How much gravitational potential energy is lost when a 0 How much gravitational potential energy is lost when a 0.25 kg baseball falls a distance of 10 m? ∆ PEg = mg ∆h = (0.25 kg)(9.8 m/s2)(-10m) = 24.5 J

The work done against gravity (and therefore the PEg gained) when raising an object to a higher level only depends on . . The force to lift the object (mg) and the vertical distance it moves NOT the path it takes to get there.

A student needs to lift a 2 kg box to the top of a 1 A student needs to lift a 2 kg box to the top of a 1.0 meter high table. The student can lift the box straight up or use a 2.0 meter long ramp. How much work does the student do (how much PEg does the box gain) . . . . to lift the box straight up? W=Fd (up) or ∆ PEg = mg ∆h 19.6 J when using the ramp?

How much force does the student use when lifting straight up? W= Fd 19.6 J = F (1.0 m) F = 19.6 N How much force does the student use when using the ramp? W = Fd 19.6 J = F (2 m) F= 9.8 N

Review Doing work on an object can lift the object to a higher vertical position. This work increases the object’s PEg no matter how it’s done. (Fd = mg ∆h) What if work is done but the object doesn’t move higher? The work can increase the object’s speed. The work done can increase the object’s Kinetic Energy.

Kinetic Energy Kinetic Energy is the energy of motion If an object speeds up or slows down, its KE changes. If its speed doesn’t increase or decrease, its KE doesn’t change. KE = ½ mv2 ∆ KE = ½ mvf2 – ½ mvi2 (not ½ m (∆ v)2 The work done (Fd) = ∆ KE

Bicyclist’s work turns into PEg, PEg turns into KE (speed increases) as bicyclist descends. Gravity does the work that increase KE.

KE = ½ mv2 KE and m are directly related. If m increases, KE increases in the same way. If m is doubled, KE is doubled KE and v are directly related. If v is doubled, KE is quadrupled.

Work = ∆ KE work done = ∆ KE Fd = ½ mvf2 – ½ mvi2 If you are driving twice as fast, how much more stopping distance would you need to stop?

Examples What is the kinetic energy of a 2000 kg car that is moving at 35 m/s? KE = ½ mv2 = ½ (2000 kg)(35 m/s)2 1.225 x 106 J

A 10 kg object has a kinetic energy of 4000 J A 10 kg object has a kinetic energy of 4000 J. How fast is the object moving? KE = ½ mv2 4000 J = ½ (10 Kg) v2 v = 28.3 m/s

A child walking with a kinetic energy of 10 joules starts running for the ice cream truck. If the child quadruples their speed, what is their new kinetic energy? KE = ½ mv2 16 x 10 J = 160 J

A car has KE of 5 x 105 J when going a certain speed A car has KE of 5 x 105 J when going a certain speed. If it slows to 1/2 its original speed, what is its KE? KE = ½ mv2 ¼ (5 x 105 J)

A bicycle moving at 3 m/s has a kinetic energy of 180 joules A bicycle moving at 3 m/s has a kinetic energy of 180 joules. The bicyclist slows down so that their kinetic energy is 20 joules. What is the bicyclist’s new speed? KE = ½ mv2 1/9 KE 1/3 speed = 1 m/s

d = v2 You are driving 30 m/s when you are forced by a deer to drastically deviate from your normal speed. You slam on the brakes, and skid to a stop in a distance of 180 m, missing the deer by half an antler. If you had been going only 15 m/s, how far would you have skidded before coming to a stop?

d = v2 A car going 30 km/h brakes, and skids 10 m to a stop. How far will this car skid if it is going 60 km/h, and then the same braking force is applied? 40 meters

d = v2 A car moving at 10 m/s is able to stop in a distance of 20 m. What would the stopping distance be if the car was moving at 40 m/s? 320 meters

How much net work is done in accelerating a 5-kg box from rest to a speed of 4.0 m/s? Work = ∆ KE = ½ mvf2 – ½ mvi2 Work = ½ (5 kg)(4 m/s)2 = 40 J

How much work must be done to accelerate a 10 kg block from a speed of 2.5 m/s to a speed of 8.0 m/s? Work = ∆ KE = ½ mvf2 – ½ mvi2 Work = ½ (10 kg)(8 m/s)2 - ½ (10 kg)(2.5 m/s)2 = 320 J - 31.25 J = 289 J

An 5 kg object is dropped from the top of a 15 meter high building An 5 kg object is dropped from the top of a 15 meter high building. How fast is the object moving when it hits the ground? Now we have 2 ways to find this out.

Review Work can be done on or by something by being lifted or by falling. Work (Fd) = ∆ PEg = mg ∆h Work can be done on or by something when it speeds up or slows down. Work (Fd) = ∆ KE = ½ mvf2 – ½ mvi2 Work can be done on or by a spring when it is stretched or compressed.

Elastic Potential Energy When work is done to stretch or compress a spring (or anything elastic) it gains PEs. PEs = ½ kx2 Work done (Fd) = ∆ PEs ∆ PEs = ½ kxf2 – ½ kxi2 not ∆ PEs = ½ k (∆ x)2 k = spring constant (# that describes the stiffness of that spring, higher k = stiffer spring)

X is how far spring is stretched from the equilibrium position.

Hooke’s Law What is the relationship between the force applied to a spring (F) and how far it stretches (x)? x = F (directly related) one increases so does the other.

What is the relationship between the spring constant (k) and how far it stretches (x)? X= 1/k (inversely related) higher k (stiffer spring), less stretch x = F k F = k x

Doubling the spring constant, would half the stretch, given the same force. Doubling the force, would double the stretch, given the same spring

To use W = Fd to solve for ∆ PEs, you must use the average force. If you don’t have the average force (you just know F) you must use F = k x, solve for k and then plug into ∆ PEs = ½ kxf2 – ½ kxi2

Examples A force of 30 N is necessary to compress a spring 1.2 m from its equilibrium position. What is the spring constant of this spring? F = k x 30 N = k (1.2 m) K = 25 N/m

A spring has a spring constant of 10 N/m A spring has a spring constant of 10 N/m. What is the elongation of the spring if a force of 5 N is used to stretch the spring? F = k x 5 N = 10N/m (x) x = 0.5

What is the potential energy of a spring, with a spring constant of 60 N/m, if it is stretched 0.3 m from its equilibrium position? ∆ PEs = ½ kxf2 – ½ kxi2 = ½ 60 N/m (0.3m)2 – 0 = 2.7 Nm 2.7 J of work is done on the spring

A student does 2.0 joules of work in stretching a spring beyond its equilibrium position. If the spring has a spring constant of 20 N/m, how far did the student stretch the spring? ∆ PEs = ½ kxf2 – ½ kxi2 2 J = ½ (20 N/m)(x)2 – 0 0.45 m

A force of 100 N is used to stretch a spring 2 meters A force of 100 N is used to stretch a spring 2 meters. How much elastic potential energy is stored in the spring? W = Fd ? Don’t have average force, so . . . . F = k x 100 N = k (2 m) K = 50 N/m

∆ PEs = ½ kxf2 – ½ kxi2 = ½ (50 N/m)(2 m)2 100 J You would have to solve for work this way also.

An average force of 200 N is used to stretch a spring a distance of 0 An average force of 200 N is used to stretch a spring a distance of 0.25 meters. How much elastic potential energy does the spring have? W = F d = (200 N)(0.25 m) = 50 J You MUST solve for PES and W this way