Physics 1D03 - Lecture 19 Review of scalar product of vectors Work by a constant force Work by a varying force Example: a spring Work and Energy

Physics 1D03 - Lecture 19 Work and Energy Energy approach: Net work = increase in kinetic energy - acceleration at any instant is caused by forces - equivalent to Newton’s dynamics - scalars, not vectors - compares energies “before and after” Newton’s approach:

Physics 1D03 - Lecture 19  The scalar product or dot product of two vectors gives a scalar result: vector vector = scalar Math Review

Physics 1D03 - Lecture 19 ScaIar product and cartesian components: (note the right-hand-side is a single scalar) To prove this, expand using the laws of arithmetic (distributive, commutative), and notice that since, i, j, k are mutually perpendicular since they are unit vectors and

Physics 1D03 - Lecture 19 MATH QUIZ A constant force is applied to an object while it undergoes a displacement The work done by is :

Physics 1D03 - Lecture 19 Work   Work by a constant force F during a displacement s: Units : N m = joule (J) Work = (component of F parallel to motion) x (distance) We can also write this as: 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) Work = F s = Fscos(θ)

Physics 1D03 - Lecture 19 Example (massless pulleys, no friction) s = 2 m How much work is done on the rope by F p ? How much work is done by the upward force on the ball? 100 N FPFP

Physics 1D03 - Lecture 19 Quiz The two forces, P and F g are constant as the block moves up the ramp. The total work done by these two forces combined is: 3 m 2 m a)20 J b) c)40 J F g = 5 N P = 10 N

Physics 1D03 - Lecture 19 Quiz m f k = 50 N w = 100 N n F p = 120 The block is dragged 2.5 m along the slope. Which forces do positive work? negative work? zero work?

Physics 1D03 - Lecture 19 Quiz m f k = 50 N w = 100 N n F p = 120 The block is dragged 2.5 m along the slope. Find : a)work done by F p b)work done by f k c)work done by gravity d)work done by normal force e)Total work on the block

Physics 1D03 - Lecture 19 a)W p = (120 N)(2.5 m) = 300 J b)W f = (- 50 N)(2.5 m) = -125 J c)W g = d)W n = 0 (  motion) Total : (- 125) + (- 150) = 25 J mg s

Physics 1D03 - Lecture 19 Example: How much work is done to stretch a spring scale from zero to the 20-N mark (a distance of 10 cm)? We can’t just multiply “force times distance” because the force changes during the motion. Our definition of “work” is not complete. Forces which are not constant: Varying force: split displacement into short segments over which F is nearly constant. F(x)  x F For each small displacement  x, the work done is approximately F(x)  x, which is the area of the rectangle. x

Physics 1D03 - Lecture 19 Work is the area (A) under a graph of force vs. distance Split displacement into short steps  x over which F is nearly constant... F(x) x x Take the limit as  x  0 and the number of steps   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. A

Physics 1D03 - Lecture 19 In 1D (motion along the x-axis): 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 In the limit as  x goes to zero,

Physics 1D03 - Lecture 19 Example: An Ideal Spring. Hooke’s 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 FsFs F s  kx Find a function W(x) so that

Physics 1D03 - Lecture 19 Concept 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)W b)2W c)3W d)4W