Lesson 27 – Work Done by a Constant Force

1. Lesson 27 – Work Done by a Constant Force By: Fernando Morales September 05, 2013

2. Learning Goals • Develop a basic understanding of the different types of systems • Develop a definition of work • Investigate cases where the work is done on the system vs. by the system vs. no work

3. Classification of Systems • Open System – An open system can exchange both matter and energy with its surroundings. • Closed System – Matter does not enter or leave a closed system, but energy can enter or leave. • Isolated System – Neither matter nor energy can enter or leave an isolated system.

4. Energy and Work • Energy is the ability to do work • Work is the ability to transfer energy • The law of conservation of energy states that, in an isolated system, the total energy is conserved, but can be transformed from one form to another • In other words, energy is neither created nor destroyed, it merely changes from one form to another

5. How do we multiply vectors? • Work is a scalar quantity • F is the applied force (not the net force!) • d is the displacement • Work is the result of a vector multiplication known as the dot product

6. Dot Product • The dot product of two vectors is defined as follows • Only the component of the force in the same direction as the displacement does work

7. X6 - 0 Work examples Find the work done EXAMPLE Solution F = 10 N θ = 30º W = Fd cos 30º = 43.3 J d = 5 m F = 20 N θ = 135º W = Fd cos 135º = -354 J d = 25 m F = 30 N * Watch out – check θ – you need the angle between F and d θ = 30º * W = Fd cos 120º = -75 J d = 5 m

8. Work done by a constant force We define workto be the product of displacement and the component of the force parallel (||) to the displacement W = F|| d W = F d cosθ Work comes in Joules 1 J = 1 Nm = 1 kg m²/ s²

9. Peer Instruction • Can a centripetal force ever do work? Explain [T] • Answer: A centripetal force is always perpendicular to the direction of motion; work is only done by forces in the direction of motion. Thus a centripetal force never does any work on an object.

10. Work is a transfer of energy The work done on the nail is +ve … Wn = Fd > 0 … The work done on the hammer is –ve … Wh = -Fd < 0 … Work has been done onthe nail bythe hammer Energy was transferred fromthe hammer tothe nail A hammer hits a nail … exerting a force F on the nail … … but it doesn’t end there; the energy is transferred from the nail to the wood by friction and dissipated as heat, sound energy. by the 3rd law … the nail exerts force –F on the hammer …

11. Peer Instruction • If no work is done when I push on a wall, then why do I find it tiring? [C] • Answer: While pushing, my muscles are tensed, and energy is expended maintaining this tension. It is the expenditure of energy which we find tiring, not necessarily the work done with the energy. As well, psychologically, pushing a wall is boring.

12.  Work = 0 work done by a force which is  to the direction of the motion = 0 cos  = 0 when  = 90° Work is done only by a force acting in the direction of motion Man does no work on bag: force is perpendicular to displacement

13. Peer Instruction • Can a normal force ever do work? Explain [T] • Answer: If the object moves in the direction of the normal force then the normal force does work. For example, if you push a box across the floor, the force between you and the box is normal (or at least has a normal component) and it does work. • Or imagine an elevator with a box on the floor; as the elevator rises the box also rises – the forces on the box are force of gravity down and the normal (due to the elevator floor) so the normal must do work.

14. Net Work done Net Work is done by the Wnet = W = F·d = Fnet d … Net Force ! Net work is the sum of the work done by the individual forces X6 - 1 Work done on a crate A 50 kg crate is pulled 40 m along a horizontal floor by a constant force exerted by a person, FP = 100 N acting at 37º. The floor is rough, Ffr = 50 N. Find the work done by each force and the net work done on the crate. EXAMPLE Solution There are 4 forces to consider WGravity = mg x cos 90º = 0 WNormal = FN x cos 90º = 0 WP = FP x cos 37º = 3200 J Wfr = Ffr x cos 180º = -2000 J Note: Fnet = 30 N || to x so Wnet = Fnet· x = 1200 J Wnet = 3200 J – 2000 J = 1200 J

15. Work on a backpack (a) find the work a hiker does carrying a 15.0 kg back pack up a 10.0 m high hill (b) the work done by gravity on the pack and (c) the net work done on the pack (assume the motion is smooth and at constant velocity) EXAMPLE m s² X6 - 3 Solution We ignore any horizontal forces (motion is considered smooth); there are 2 forces Fy = may FH – mg = 0 FH = mg = (15.0 kg)(9.81 ) =147 N (a) WH = FH(d cos θ) dcos θ = h  WH = mgh = 1470 J (b) WG = FG d cos (180º - θ) cos (180º - θ) = - cos θ  WG = FG d (- cos θ)  WG = -mgh = -1470 J (c) The net work is WG + WH = 0 Not surprising as the net force on the backpack was 0

16. Required Before Next Class • Stay Tuned on the Website, did not have enough time to go over the questions on textbook