Work and Energy

1. Work and Energy Physics 1D03 - Lecture 19

2. Work   Work by a constant force F during a displacement s: Work = F• s = Fscos(θ) Units : N • m = joule (J) Physics 1D03 - Lecture 19

3. Example 2.5 m Fp = 120 w= 100 N 5 3 fk = 50 N 4 n The block is dragged 2.5 m along the slope. 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

4. Wp = (120 N)(2.5 m) = 300 J • Wf = (- 50 N)(2.5 m) = -125 J • Wg = • Wn = 0 ( motion) s mg Total : 300 + (- 125) + (- 150) = 25 J Physics 1D03 - Lecture 19

5. 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

6. 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   Physics 1D03 - Lecture 19

7. 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

8. F(N) 5 x(m) 0 1 2 3 4 5 6 Determine the work done by a force as the particle moves from x=0 to x=6m: Variable Force: Physics 1D03 - Lecture 19

9. 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 and E=½kx2 Physics 1D03 - Lecture 19

10. Example: Work by a Spring Fs Fs= - kx and Es=½kx2 When x=0, Energy=0, when sprig is at x=A, thenE=½kA2. Physics 1D03 - Lecture 19

11. Concept 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

12. Concept 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

13. Kinetic Energy Definition: for a particle moving with speed v, the kinetic energy is K = ½ mv2 (a SCALAR) Then the Work-Energy Theorem says: The total work done by all external forces acting on a particle is equal to the increase in its kinetic energy. Physics 1D03 - Lecture 19

14. Example A bartender slides a 1-kg glass 3 m along the bar to a customer. The glass is moving at 4 m/s when the bartender lets go, and at 2 m/s when the customer catches it. Find the work done by friction, and calculate the force of friction. Physics 1D03 - Lecture 19