# Work. Work = Force x parallel distance (parallel component of displacement) F v

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1 Work Work = orce x parallel distance (parallel component of displacement) W k = d parallel d parallel Units: N m= J = " joules" = ( kg m2/ s2) = average force computed over the distance r r When is not parallel to r d, then we must take the component of which is parallel to d. θ v θ v h h = cosθ h v = sinθ W = d= (cos θ) d k h d If = 100 N and θ = 30 degrees, Compute v and h If d = 5 m, compute W * for those of you who have had advanced math, the parallel component is computed by the dot product of the force and displacement vectors.

2 Example: If you pull on a wagon with a force of 100 N at an angle of 30 degrees w.r.t. the horizontal and you pull over a distance of 5 m, how much work do you do on the wagon? Know: = 100 N; d = 5 m; θ = 30 Need: W k Use: W k = ( d) parallel = cosθ d W k = 100 N cos 30 5 m = 433 N m Positive v. Negative Work If and d parallel point in the same direction the work done is positive d If and d parallel point in opposite directions the work done is negative d

3 Energy = the capacity to do work (scalar) Types of energy: mechanical, chemical, heat, sound, light, etc We are most interested in mechanical energy Mechanical Energy Kinetic Energy (KE) - energy due to motion KE = 1 mv2 unit: kg m2/ s2= J 2 e.g. a diver (mass = 70 kg) hits the water after a dive from the 10 m tower with a velocity of 14 m/s. How much KE does she possess? KE = 1mv2 = 170 kg( 14 m/ s) 2 = 6860 J 2 2

4 Mechanical Energy (Gravitational) Potential Energy (P.E.) energy due to the change of position in gravitational field PE = mgh h = height of something above some reference line m = mass g = acceleration due to gravity ( 9.8 m/s 2 ) TOTAL (J) m = 1 kg v i = 6.3 m/s = 29.4 J = 29.4 J = 29.4 J = 29.4 J = 29.4 J

5 PE > 0 PE = m( g)(+h) PE = +mgh PE < 0 PE = m( g)( h) PE = mgh +h h Units: kg m2/ s2= J The negative sign in the PE equation is necessary to account for the direction of gravity (PE > 0 when h > 0 and PE < 0 when h < 0) Potential Energy: Note: in the absence of air resistance and other resistive forces - - PE can be completely converted to KE by the work done by gravity on the way down. e.g. a diver on top of a 10 m tower has a positive PE compared to water level PE = mgh = 60 kg ( 9.8 m/s 2 ) (+10 m) = 6860 J NOTE : This value is identical to that found in the kinetic energy example.

6 Mechanical Energy Strain or elastic energy (SE) energy due to deformation this type of energy arises in compressed springs, squashed balls ready to rebound, stretched tendons inside the body, and other deformable structures SE = 1kx 2 2 x= amountof deformation k = stiffnessof spring( orother structure)

7 elongated (+x) resting length ( x) compressed Work-Energy Relationship The work done by the net force acting on a body is equal to the change in the body s kinetic energy ( Σd ) = 1mv 1mv parallel f i 2 2 ΣW = KE k 2 2 This relationship is true as long as there is no change in vertical position.

8 W GR v Vi = 0 W GR v Vf = 0 if KE i = 0 and KE f = 0 then KE = 0; therefore the work done by the person (GR v ) is completely offset by the negative work done by gravity (W). Overall -- no work was done because there was no change in KE overall Tell that to your muscles!!! W Alternate formulation of Work-Energy Relationship (Hay, 1993) W v Vi = 0 Vf = 0 Consider the work done by a single force. In this example consider v v v d v = KE+ PE ( + SE ) KE = 0 PE = mgh = 100kg ( 98. m/ s 2 ) 2m = 1960J This is numerically opposite to the work done by gravity. v did J of work W did 1960 J of work

9 Another example: Team A Team B 1,000,000 N +1,000,001 N d = 1 m Σ = 1,000,000 N + 1,000,001 N = 1N W k = Σ d = 1N 1m = 1Nm Example problems: Work A 60 diver leaves the 10 m board with zero vertical velocity. Compute the potential and kinetic energies at 10 m, 5 m and 0 m above the water. Compute the work done on the diver upon entry. A baseball catcher gives to stop a 22 m/s fastball over a distance of 0.25 m. What was the average force needed to stop the 0.2 kg baseball? What is the average force needed to stop the same 22 m/s fastball over m?

10 Example problems: Work A 60 kg diver leaves the 10 m board with zero vertical velocity. Know: Need: Use: Example problems: Work Distance (m) PE (J) KE (J) Total Energy At rest Work done =

11 Example problems: Work A baseball catcher gives to stop a 22 m/s fastball over a distance of 0.25 m. Know: Need: Use: Power the rate at which work is done the rate at which energy is expended P = W k = work t time units: J joule s = s = watt ( W)

12 Example if m = 100 kg g = 9.8 m/s 2 h = 2 m W k = mgh = 100(9.8)(2) = 1960 J now add time Case 1: raise the barbell slowly -- t = 5 s P = W t k = 1960 J / 5 s = 392 W Case 2: raise the barbell quickly -- t = 1.5 s P = W t k = 1960 J / 1.5 s = W Power - alternate form of equation P W d k parallel = = = v t parallel t Power can be θ positive or negative v depending on whether P = cosθ v and v point in same general direction (+ power) or in opposite directions ( power)

13 Example problems: Work & Power Arizona State University football coach Bruce Snyder banished 242 lb (1078 N) Terry Battle and 205 lb (913 N) Jake The Snake Plummer to the Sun Devil Stadium steps for some corrective therapy following a poor practice. The 50 stadium steps require a vertical rise of 15 m. Battle completed 8 trips up the stairs in 195 s and Plummer completed 12 trips in 270 s. Which athlete performed more work? Which athlete had the higher average power output? (Ignore the return trips down the steps and any horizontal displacement) Example problems: Work & Power Two soccer players work out in the off season by sprinting up a 40º hill for a distance of 100 m before stopping, resting, and walking back down. Mia has a mass of 60 kg. Julie has a mass of 65 kg. It takes Mia 42 sec. and Julie 38 sec. to reach the top of the hill. Which athlete performed more work? Which athlete had the higher average power output? (Ignore the return trips down the steps and any horizontal displacement)

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### Roller Coaster Mania!

Overview Roller Coaster Mania! This series of educational programs was designed to simultaneously entertain and challenge gifted youth in their time outside of the school setting; however, the activities

### v v ax v a x a v a v = = = Since F = ma, it follows that a = F/m. The mass of the arrow is unchanged, and ( )

Week 3 homework IMPORTANT NOTE ABOUT WEBASSIGN: In the WebAssign versions of these problems, various details have been changed, so that the answers will come out differently. The method to find the solution

### W i f(x i ) x. i=1. f(x i ) x = i=1

Work Force If an object is moving in a straight line with position function s(t), then the force F on the object at time t is the product of the mass of the object times its acceleration. F = m d2 s dt