Test TODAY at 9:30 am in CNH-104 !!! Physics 1D03 - Lecture 19

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

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

Math Review The scalar product or dot product of two vectors gives a scalar result: vector • vector = scalar  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 and since, i, j, k are mutually perpendicular since they are parallel unit vectors Physics 1D03 - Lecture 19

Work   Work by a constant force F during a displacement s: Work = (component of F parallel to motion) x (distance) We can be write this as: Work = F • d = Fdcos(θ) 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

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

What if the person were to walk up the stairs ? Example A person walks down the stairs. Is the work done by gravity positive or negative ? What if the person were to walk up the stairs ? Physics 1D03 - Lecture 19

Example You are pulling a sled with a force of 50N at an angle of 30o to the horizontal, for a distance of 25m. Determine the work done by you in pulling the sled. Physics 1D03 - Lecture 19

Quiz Fg = 5 N P = 10 N 2 m 3 m 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 b) c) 40 J Physics 1D03 - Lecture 19

Example 2.5 m Fp = 120 w = 100 N fk = 50 N n A block is dragged 2.5 m up along the slope. Which forces do positive work? A) negative work? B) zero work? Physics 1D03 - Lecture 19

The block is dragged 2.5 m up along the slope of 37o. Find : Example 2.5 m Fp = 120 w = 100 N fk = 50 N n The block is dragged 2.5 m up along the slope of 37o. Find : work done by Fp work done by fk work done by gravity work done by normal force Total work on the block Physics 1D03 - Lecture 19

Total : 300 + (- 125) + (- 150) = 25 J Wp = (120 N)(2.5 m) = 300 J Wf = (- 50 N)(2.5 m) = -125 J Wg = Wn = 0 ( motion) 2.5m d=2.5m sin(37o) F=mg Total : 300 + (- 125) + (- 150) = 25 J Physics 1D03 - Lecture 19

5 min rest 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 can’t 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) x For each small displacement Dx, the work done is approximately F(x) Dx, which is the area of the rectangle. F D x Physics 1D03 - Lecture 19

Work is the area (A) under a graph of force vs. distance We get the total work by adding up the work done in all the small steps. As we let Dx become small, this becomes the area under the curve, and the sum becomes an integral. F(x) F(x) A x x Split displacement into short steps Dx over which F is nearly constant... Take the limit as Dx 0 and the number of steps   Work is the area (A) under a graph of force vs. distance Physics 1D03 - Lecture 19

Variable Force: Determine the work done by a force as the particle moves from x=0 to x=6m: x(m) F(N) 0 1 2 3 4 5 6 5 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 Dx is, approximately, DW  F(x) Dx. Rearrange to get In the limit as Dx goes to zero, Physics 1D03 - Lecture 19

An Ideal Spring Hooke’s Law: The tension in a spring is proportional to the distance stretched. or, F = kx 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 and E=½kx2 Physics 1D03 - Lecture 19

Show that for a spring, E= ½ kx2. Example: Show that for a spring, E= ½ kx2. Physics 1D03 - Lecture 19

Quiz A spring is hanging vertically. A student attaches a 0.100-kg mass to the end, and releases it from rest. The mass falls 50 cm, stretching the spring, before stopping and bouncing back. During the 50-cm descent, the total work done on the mass was: zero 0.49 J -0.49 J none of the above Physics 1D03 - Lecture 19

Quiz A physicist uses a spring cannon to shoot a ball at a 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: W 2W 3W 4W Physics 1D03 - Lecture 19