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

2. 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

3. 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

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

5. 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

6. 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

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

8. 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

9. 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

10. 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

11. 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

12. 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

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

14. 5 min rest
Physics 1D03 - Lecture 19

15. 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

16. 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

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

18. 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

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

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

21. Quiz
A spring is hanging vertically. A student attaches a 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

22. 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