Work 1 Work done by a constant force

Work 1. Work done by a constant force: gravity 2. Work done by a variable force: a spring 3. Work done by a variable force: general case Sections 7. 6 – 7. 9

Work-kinetic energy theorem The theorem says that the change in kinetic energy of a particle is the net work done on the particle. It holds for both positive and negative work: If the net work done on a particle is positive, then the particle’s kinetic energy increases by the amount of the work, and the converse is also true.

Work-Energy Theorem: Question § A 2 kg box slides over a frictionless floor § F 1 = 10 N, F 2 = 2 N § The initial velocity of the box is v 0=3 m/s § The box slides 4 m in the +x direction § Q: Rank each situation for its final kinetic energy, greatest to least a) b) c) d) e) c, b, d, a d, b, c, a a, b, c, d a, d, b, c a, c, b, d 3

Solution = E 2/22/2021 4

Solution W = Kf – Ki § Kf = ½mvi 2 + net work done by forces on box = ½(2 kg)(3 m/s)2 + Wnet = 9 J + Wnet (a) Kf = 9 J + [(10 N)cos 30 o + (2 N)](4 m) = 52 J (b) Kf = 9 J + [(10 N)cos 30 o - (2 N)](4 m) = 36 J (c) Kf = 9 J + (10 N)(4 m) = 49 J (d) Kf = 9 J + [(10 N)sin 30 o - (2 N) sin 30 o](4 m) = 25 J Answer: a, c, b, d 2/22/2021 Lecture 11 – Work 5

Work Done by a Constant Force (Gravity) § An object is thrown upward and decelerates due to gravity. § Work done by the gravitational force Wg = mgdcosf = mgdcos 180 o = -mgd § Why a minus sign? § Because F and d are oppositely directed. § As the object heads downward what is the work? Wg = mgd cos 0 o = mgd, § Positive work done by the gravitational force on the tomato (F & d parallel). 2/22/2021 6

Work Done by the Gravitational Force (a) An applied force lifts an object The object’s displacement makes an angle f =180° with the gravitational force on the object Wg = mgd cos 180 o = - mgd The applied force does positive work on the object. Wa = mgd cos 0 o =mgd (b) An applied force lowers an object. The displacement of the object makes an angle with the gravitational force. Wg = mgd cos 0 o =mgd The applied force does negative work on the object

The work done by gravity during the descent of a projectile is: A) positive B) negative C) zero D) depends for its sign on the direction of the y axis E) depends for its sign on the direction of both the x and y axes 2/22/2021 8

Work Done by a Variable Force (Spring) § Forces on a spring – Relaxed state (equilibrium) – Stretched: restoring force, F < 0 – Compressed: restoring force, F > 0 § Empirical law: Hooke’s Law (massless, ideal spring) § In 1 -dimension, F = - k x 2/22/2021 Lecture 12 – Work and Power 9

Work Done by a Spring § k is called the spring constant § The spring constant measures the stiffness of the spring § k has units of N/m. § Hooke’s Law, F(x) = -kx, is a variable force The force depends on displacement in a linear manner 2/22/2021 Lecture 12 – Work and Power 10

Work Done by a Spring § Since the force applied by the spring is not constant, how do we find the work? § We know W = F x is true when F is constant, so we do the following: – Divide interval into small segments § Small enough so in Dx, F is constant! Magnitude of F only shown here. – Find the work in each segment – Add them up! W = S(Fi. Dx) 2/22/2021 Lecture 12 – Work and Power 11

Work Done by a Spring § You can calculate the area under the line by considering it as a rectangle plus a triangle. (Consider only the magnitude of F for now): Ff Magnitude of F only shown here. Area = (xf - xi) Fi + ½(xf - xi) (Ff - Fi)= Fi = (xf - xi) k xi + ½(xf - xi) (k xf – k xi) = ½kxf 2 – ½kxi 2. If you recognize that F=-k x, this needs to be Reversed: Work = ½ k xi 2 – ½ k xf 2 2/22/2021 Lecture 12 – Work and Power xf 12

Work Done by a Spring § What is the more sophisticated method for adding under a curve? Integration ! § Integrating the force from the initial to final position of the mass gives us Work done by a spring force: 2/22/2021 Lecture 12 – Work and Power Initial term first! 13

Work Done by a Spring Force Hooke’s Law: To a good approximation for many springs, the force from a spring is proportional to the displacement of the free end from its position when the spring is in the relaxed state. The spring force is given by The minus sign indicates that the direction of the spring force is always opposite the direction of the displacement of the spring’s free end. The constant k is called the spring constant (or force constant) and is a measure of the stiffness of the spring. The net work Ws done by a spring, when it has a distortion from xi to xf , is: Work Ws is positive if the block ends up closer to the relaxed position (x =0) than it was initially. It is negative if the block ends up farther away from x =0. It is zero if the block ends up at the same distance from x= 0.

Sample Problem: Work done by spring

Work Done by a Force Applied to the Block § Apply a force, Fa , to a block which has a spring connected on one side. The force results in compression of the spring. § As you do work Wa on the block, the spring does work, Ws, on the block § Work-KE theorem: Kf – Ki = Wa + Ws § If the block is stationary before and after the displacement then Kf – Ki = 0, so Wa = - Ws § What does this mean physically? 2/22/2021 Lecture 12, Work and Power 16

Work Done by an Applied Force § Work applied to block is equal and opposite to the work done by the spring § If the block is not stationary after the displacement, is this statement true? § NO!, since Ws = Kf – Ki – Wa the work done by the spring would also depend on DK 2/22/2021 Lecture 12, Work and Power 17

Sign Convention: work done by spring force The work done BY the spring is W = ½k(xi 2 – xf 2 ) § W>0 – is true when |xi| > |xf |, and the block ends up closer to the equilibrium position. – means the spring transfers energy to the block. § W<0 – is true when |xi| < |xf |, and the block ends up farther from the equilibrium position then where it started. – means the block transfers energy to the spring. – Think of a hand pulling the block farther from the equilibrium position (Wa = -Ws). 2/22/2021 Lecture 12, Work and Power 18

Sign Convention: work done by spring force W = ½ k (x i 2 – x f 2 ) § W=0 – is true when |xi| = |xf |, and the block ends up at the same distance from the equilibrium position it started at – means no energy is transferred 2/22/2021 Lecture 12, Work and Power 19

Sign Convention: work done by spring force W = ½ k (x i 2 – x f 2 ) § What is the physical meaning of the absolute value signs, |x|, for each of these situations? – Consider W = 0. The block ends up at the same distance from the equilibrium point, but it could be physically on the opposite side from whence it started – Since work depends on the square of position, the distance from the equilibrium point can be determined, but not the displacement vector! 2/22/2021 Lecture 12, Work and Power 20

Summary of the Spring § The block shown in the figure is stationary at a given xi and moved to a final position xf, where it is again stationary § Given the following initial and final displacement (see below), is the work done by the spring on the block +, - or 0? xi, xf = – 3 cm, 2 cm (a) + (b) 0 (c) 2/22/2021 Lecture 12, Work and Power 21

Answer -3 cm, 2 cm (a) Ws > 0. Initially the block is farther away so, |xi| > |xf | and the work done by the spring is positive. 2/22/2021 Lecture 12, Work and Power 22

Summary of the Spring § § The block shown in the figure is stationary at a given xi and moved to a final position xf, where it is again stationary. Given the following initial and final displacement is the work done by the spring on the block +, - or 0? xi, xf = – 2 cm, 2 cm (a) (b) (c) 2/22/2021 + 0 Lecture 12, Work and Power 23

Answer -2 cm, 2 cm (b) Ws = 0. The block begins and ends at the same distance from the equilibrium point. 2/22/2021 Lecture 12, Work and Power 24

General Variable Force One dimensional case: We can make the approximation better by reducing the strip width Dx and using more strips (Fig. c). In the limit, the strip width approaches zero, the number of strips then becomes infinitely large and we have, as an exact result: 2/22/2021 Lecture 12, Work and Power 25

General Variable Force B. Three dimensional force: If where Fx is the x-components of F and so on, and where dx is the x-component of the displacement vector dr and so on, then Finally,

Work-Kinetic Energy Theorem A particle of mass m is moving along an x axis and acted on by a net force F(x) that is directed along that axis. The work done on the particle by this force as the particle moves from position xi to position xf is : But, Therefore,

Power § Power is the rate at which work is done § The average power is § The instantaneous power is 2/22/2021 Lecture 12, Work and Power 28

Power Units § Power is expressed in Watts 1 Watt = 1 W = 1 J/s § Horsepower (hp) is also a power unit 1 hp = 746 W § Work is sometimes expressed as the product of power and time (kilowatt-hour, k. W∙h) 1 k. W∙h = (1000 W)(3600 s) = 3. 6 MJ (1 MJ = 106 J) 2/22/2021 Lecture 12, Work and Power 29

While you sleep, your body is using energy at a rate of P = 77 Watt. How many kilocalories (kcal) are used during an 8 hour period? 1 kcal is equal to 4186 joules. a) 66 kcal b) 240 kcal c) 530 kcal d) 710 kcal e) 1200 kcal

While you sleep, your body is using energy at a rate of P=77 Watt. How many kilocalories are used during an 8 hour period? 1 kcal is equal to 4186 joules. a) 66 kcal b) 240 kcal c) 530 kcal d) 710 kcal e) 1200 kcal

Power and Force § If a constant force is applied to an object moving in a straight line which is the scalar product, 2/22/2021 Lecture 12, Work and Power 32

Example of Power Calculation § A 1000 kg car requires 12 hp from its engine to cruise at a steady speed of 80 km/h on a flat road § How much power would it require to move up a 10 o incline (~ 18% grade) at the same constant speed? (Assume the total frictional force is that due to the tires on the road (e. g. improperly inflated), the bearings, friction in the motor, etc. , and is denoted by f. R, plus there is a constant air drag force, f. D. Remember: Fnet can = 0 (v constant), but individual forces can still provide (or require) power. 2/22/2021 Lecture 12, Work and Power 33

Example of Power Calculation f. R + f. D +x F § On the level road, F – (f. R + f. D) = ma = 0 since the speed is constant, so F = f. R + f. D § Here the power due to the constant force, F, causes the car to move forward with constant velocity. § Since P = Fv, F=(12 hp)x(746 W/hp) / [(80 km/hr)/(3. 6 m·hr/km·s)]= 400 N – Remember to convert P in hp to W and v to m/s! N F´ § On the incline the FBD is f. R + f. D q 2/22/2021 Lecture 12, Work and Power mg 34

Example of Power Calculation § The force to move up the incline, F´, is resisted by friction, air drag plus the component of gravity along the incline. § Since the velocity is constant: F´ - (f. R + f. D) - mgsinq = 0 N F´ f. R + f. D q mg F´ = F + mgsinq where we have used our previous result that F = f. R + f. D 2/22/2021 Lecture 12, Work and Power 35

Example of Power Calculation N § From P´ = F´ v we obtain P´ = 63 hp for a speed of 80 km/h, an increase of over 5 times compared to the power required on a level road f. R + f. D § Does the friction between the road and tires change on q the incline, since the normal force changes? mg F´ § Not much, since mgcos(10 o) =. 985 mg ≈ mg and we can overlook that “detail” (just for now) – in fact friction when driving a car does not relate just to the normal force anyway - friction in the bearings, axles and motor, and air drag, are dependent on factors other than N. 2/22/2021 Lecture 12, Work and Power 36