Work ! Review of scalar product of vectors Work by a constant force Work by a varying force Example: a spring Serway & Jewett 7.1 7.3 Physics 1D03 - Lecture 19 Work and Energy Newtons approach: F ma

- acceleration at any instant is caused by forces Energy approach: Net work = increase in kinetic energy - equivalent to Newtons dynamics - scalars, not vectors - compares energies before and after Physics 1D03 - Lecture 19 B Math Review The scalar product or dot product of two vectors gives a scalar result: A vector vector = scalar

A B (magnitude of A) x (component of B parallel to A) ABcos( ) Physics 1D03 - Lecture 19 ScaIar product and cartesian components: Ay j Az k A A i x B Bx i By j Bz k Then, A B Ax Bx Ay By Az Bz (note the right-hand-side is a single scalar)

To prove this, expand A B using the laws of arithmetic (distributive, commutative), and notice that i j j k i k 0 since, i, j, k are mutually perpendicular and i i j j k k 1 since they are unit vectors Physics 1D03 - Lecture 19 Work F F FII

s Work by a constant force F during a displacement s: Work = (component of F parallel to motion) x (distance) We can also write this as: Work = F s = Fscos() Units : N m = joule (J) This is the scalar product, or dot product. Work is a scalar. If work is done on a system, W is positive (eg: lifting an object). If work is done by a system, W is negative (eg: object falling) Physics 1D03 - Lecture 19 MATH QUIZ A constant force F (1i 2 j 3k ) N is applied to an

while it undergoes a displacement object s (2i 2 j 2k ) m. F The work done by is : a) (2i 4 j 6k ) J b) 22 4 2 6 2 J c) 12 J Physics 1D03 - Lecture 19 Example (massless pulleys, no friction) FP s=2m

100 N How much work is done on the rope by Fp? How much work is done by the upward force on the ball? Physics 1D03 - Lecture 19 Quiz Fg = 5 N P = 10 N 2m 3m The two forces, P and Fg are constant as the block moves up the ramp. The total work done by these two forces combined is: a) 20 J 30 2 10 2 J 32 J c) 40 J b) Physics 1D03 - Lecture 19

Question Fp = 120 w = 100 N 5 fk = 50 N m 5 . 2 3 4 n The block is dragged 2.5 m along the slope. Which forces do positive work? negative work?

zero work? Physics 1D03 - Lecture 19 Example Fp = 120 w = 100 N 5 fk = 50 N m 5 . 2 3 4 n The block is dragged 2.5 m along the slope. Find :

a) work done by Fp b) work done by fk c) work done by gravity d) work done by normal force e) Total work on the block Physics 1D03 - Lecture 19 a) Wp = (120 N)(2.5 m) = 300 J b) Wf = (- 50 N)(2.5 m) = -125 J s mg (2.5m)( 53 )(100 N) 150 J c) Wg =

d) Wn = 0 ( motion) Total : 300 + (- 125) + (- 150) = 25 J Physics 1D03 - Lecture 19 Forces which are not constant: Example: How much work is done to stretch a spring scale from zero to the 20-N mark (a distance of 10 cm)? We cant just multiply force times distance because the force changes during the motion. Our definition of work is not complete. Varying force: split displacement into short segments over which F is nearly constant. F(x)

xi xf x For each small displacement x, the work done is approximately F(x) x, which is the area of the rectangle. F x Physics 1D03 - Lecture 19 We get the total work by adding up the work done in all the small steps. As we let x become small, this becomes the area under the curve, and the sum becomes an integral. F(x) F(x)

A xi xf x Split displacement into short steps x over which F is nearly constant... W F x xi xf x Take the limit as x 0 and the number of steps

xf W F dx xi Work is the area (A) under a graph of force vs. distance Physics 1D03 - Lecture 19 xf In 1D (motion along the x-axis): W F dx xi Another way to look at it: Suppose W(x) is the total work done in moving a particle to position x. The extra work to move it an additional small distance x is, approximately, W F(x) x. Rearrange to get

W F ( x) x In the limit as x goes to zero, F ( x) dW dx Physics 1D03 - Lecture 19 Example: An Ideal Spring. Hookes Law: The tension in a spring is proportional to the distance stretched. or, |F| = k|s| The spring constant k has units of N/m Directions: The force exerted by the spring when it is

stretched in the +x direction is opposite the direction of the stretch (it is a restoring force): F = -kx Physics 1D03 - Lecture 19 Example: Work by a Spring Fs xi xf Find a function W(x) so that Fskx dW Fs ( x) dx

Physics 1D03 - Lecture 19 Quiz A physicist uses a spring cannon to shoot a ball at a stuffed gorilla. The cannon is loaded by compressing the spring 20 cm. The first 10 cm of compression requires work W. The work required for the next 10 cm (to increase the compression from 10 cm to 20 cm) would be a) b) c) d) W 2W 3W 4W Physics 1D03 - Lecture 19 Summary

For a constant force, Work = F s = (component of force parallel to motion) (distance) For a non-constant force (1-D) : Work = Fx(x) dx Physics 1D03 - Lecture 19