Energy graphs and work

Transcription

1 Energy graphs and work Saturday physics at 2pm tomorrow on music. LA info session on Monday at 5pm in UMC235 Clicker scores have been updated. If you have a 0, contact me and include your clicker ID number. Today we start Chapter 6 on work and energy. 1

2 Clicker question 1 A rider in a "Barrel O Fun" finds herself stuck with her back to the wall. Which diagram correctly shows the forces acting on her? Set frequency to BA n f s µ s mg A B C D E n No acceleration in y-direction: F net,z = f s mg = 0 In the horizontal direction: F net,r = n = ma rad 2

3 New concept: Work Applying a force over a displacement is called work. W = Fx These horses are clearly doing work. They are exerting a force over a distance 3

4 Clicker question 2 Set frequency to BA I push on a block with a force of 30 N for a distance of 2 m. How F ext = 30N much work have I done on the block? A. 0 B. 30 N C. 30 Nm D. 60 Nm E. None of the above (or impossible to tell) Work is force times displacement so W = Fx = (30 N)(2 m) = 60 Nm 2 m The unit newton meter (N m) is also known as a joule (J). 4

5 More precise definition of work Imagine a train car on a straight track so it can only move left or right. A force applied perpendicular to the way the train car can move is useless and does no work train Δ! r train However, a force applied along the direction it can move will move the train car. Work is done on the train car. In measuring work, only the part of the force that is along the displacement vector counts as work. 5

6 More precise definition of work To determine the work done by a force, we consider just the part of the force that is along the displacement vector. train θ Δ r! Using trigonometry, we can see that the component of the force along the displacement vector is Fcosθ. Thus, a more accurate description of work is W = (Fcosθ) Δr This is the definition of the dot product (also known as the scalar product)! F 6

7 Dot (scalar) product One type of vector multiplication is the dot product or scalar product which produces a scalar from two vectors C =! A! B =! A! B cosθ = ABcosθ Also, Note that!! C = A B = A B + A B + x!!! A A A = AAcos0 = A = Ax + Ay = A = Ax + Ay Az so 2 x y y where θ is the angle between the two vectors A z B z A 2 z 7

8 Dot (scalar) product The dot product can be thought of as the projection of one vector along another vector! A = ( 3,4) = 3ˆ i 4 ˆ y! j A! Let + and B = (2,0) = 2ˆ i + 0 ˆj!! C = A B = Ax Bx + AyB = = 6 y But remember, the scalar product returns a scalar, not a vector B! x 8

9 Clicker question 3 What is the dot product of the unit vector vector A! = (2,3,4)? A. 2 B. 3 C. 4 D. 5 E. 29 x x y Set frequency to BA y ˆ = (1,0,0) i z z and the! A iˆ = A i + A i + A i = = Note, this is the projection of along the vector î A! 2 Work is the dot product of the force vector and the displacement vector W =! F Δ! r =! F Δ! r cosθ = F Δr cosθ W =! F Δ! r = F x Δr x + F y Δr y + F z Δr z 9

10 More on work Work is done by a force and on an object Also, since W =! F Δ! r =! F Δ! r cosθ = F(Δr)cosθ the amount of work can be negative if the force and displacement are in opposite directions Can consider the net work on an object It is the sum of the work of all the forces and It is also the work done by the net force 10

11 Example of work A cat drags a 1 kg box at constant speed across a flat horizontal floor. The box/ floor coefficient of kinetic friction is kg box If the cat drags the box 2 m, how much work is done by the forces? Since there is no acceleration in x or y n = mg and T = µ k n = µ kmg The normal force and weight are perpendicular to the displacement and contribute no work The net work is 0 as could be seen because the net force is 11 0 v! µ k n n mg The work by the cat on the box is W Cat on box = T(Δr) = µ k mg(δr) = kg 9.8 m/s 2 2 m =12 J The work by friction is the same but negative: W Friction on box = 12 J T

12 Clicker question 4 Set frequency to BA Albert Einstein lowers a book of mass m downward a distance h at constant speed v. The work done by the force of gravity on the book is A. 0 B. mgh C. mgh D. None of the above Force and displacement are in the same direction (down) and so the work is positive and equal to force (mg) times the displacement along the force direction (h). 12

13 Clicker question 5 Set frequency to BA Albert Einstein lowers a book of mass m downward a distance h at constant speed v. The work done by the force of Einstein s hand on the book is A. 0 B. mgh C. mgh D. None of the above Force and displacement are in opposite directions and so the work is negative and equal to force (mg) times the displacement (h). 13

14 Clicker question 6 Set frequency to BA Albert Einstein lowers a book of mass m downward a distance h at constant speed v. The work done by the net force on the book is A. 0 B. mgh C. mgh D. None of the above No net force so net work done is 0. Work is done by the two individual forces but they cancel so no net work. 14

15 Clicker question 7 Set frequency to BA A 1000 kg car of mass is going around a level corner at 10 m/s. The corner has a radius of 100 m. The car travels 50 m while going around the corner. How much work is done on the car by the force of friction which is keeping the car going in a circle? Be careful here and think about what work really is. A. 0 J B. 500 J C J D J E. None of the above Remember work is the dot product of the force vector and the displacement vector. The frictional force that keeps the car in a circle is always perpendicular to the velocity so it does no work 15