Work and Energy ability to do work.
Work and Energy
Energy and Momentum provide an alternative approach to solving problems related to forces and motion.
Energy and momentum are conserved quantities, making their conservation laws valuable, especially in systems with many objects where detailed force considerations are impractical.
These laws apply to various phenomena, including the atomic and subatomic levels where Newton's laws may not be applicable.
Work and Energy are scalar quantities (no direction).
Work Done by a Constant Force
Work is defined as the product of displacement magnitude and the component of force parallel to the displacement.
Equation: W = F_{\parallel} d
W = Work done
F_{\parallel} = Component of constant force parallel to displacement
d = Magnitude of displacement
Alternative equation: W = Fd \cos{\theta}
F = Magnitude of constant force
d = Magnitude of displacement
\theta = Angle between force and displacement direction
Special Cases
Case i: Motion and force are in the same direction (\theta = 0 and \cos{\theta} = 1).
Example: Pushing a cart horizontally with 30 N force over 10 m.
W = (30 \text{ N})(10 \text{ m}) = 300 \text{ J}Define Joule (J) as the unit of work and energy.
Case ii: Motion and force are perpendicular (\theta = 90^\circ or 270^\circ, \cos{\theta} = 0).
Example: Holding a bag of groceries and walking horizontally.
No work is done as there's no force in the direction of motion.
Case iii: Motion and force are opposite (like friction, \cos{\theta} = -1).
Negative work is done by friction.
Important to specify:
Whether work is done by or on a specific object.
Whether work is done by a particular force or the net force.
Example 1
A person pulls a 50-kg crate 40 m along a horizontal floor with a constant force at a 37° angle.
The floor exerts a friction force F_{fr}.
Determine:
a) Work done by each force on the crate.
b) Net work done on the crate.
Solution:
Work done by normal force (F_N) and gravitational force (mg) is zero (right angles to displacement).
Work done by applied force (Fp): WF = F_p d \cos{\theta}
Work done by friction force (F_{fr}) is negative.
b) Net work done:
W{net} = \sum Wi = WF + W{fr}Net work is also calculated by determining the net force and its component along the displacement.
Work is negative when the force (or its component) opposes motion, causing the object to slow down.
Positive work indicates the force is speeding up the object.
Read Pg.199(1-4) and Pg.202(5-*8) from the textbook.
Example 2 (Optional)
Work on a backpack.
a) Work a hiker does on a 15.0-kg backpack to carry it up a hill of height h.
b) Work done by gravity on the backpack.
c) Net work done on the backpack (assuming smooth, constant velocity motion).
Does the earth do work on the Moon?
The Moon revolves around the earth in a nearly circular orbit, maintained by Earth's gravitational force.
Does gravity do:
a) positive work
b) negative work
c) no work on the moon?
Work Done by a Varying Force (Graphical Interpretation)
As a rocket moves away from Earth, work is done against the varying gravitational force.
Similarly, spring force varies with stretch or compression.
Work done by a varying force can be determined graphically:
Plot F_{\parallel} (= F cos θ), the component of F parallel to d, versus distance d.
Divide the distance into small segments Δd.
For each segment, indicate the average F_{\parallel} by a horizontal dashed line.
Work done in each segment is ΔW = F{\parallel}Δd, the area of a rectangle (Δd wide and F{\parallel} high).
Total work is the sum of these areas.
More segments (smaller Δd) increase accuracy, approaching the area under the curve as Δd approaches zero.
Bottom Line: Work done by a variable/constant force between two points equals the area under the F_{\parallel} vs. d curve between those points.
Example
A 24 kg rocket car is initially at rest on a frictionless horizontal surface.
The engine is ignited and the graph shows thrust force, F, versus distance travelled, d, for the rocket car.
Find the rocket car's speed after it has travelled 200 m.
\text{area bounded by graph and d - axis} = \frac{1}{2} \cdot \text{base} \cdot \text{height}
= \frac{1}{2} \times 200 \times 320
= 32 000 \text{ J}
Work-energy Theorem: W = \Delta KE
\Delta KE = KE2 - KE1 = 32 000
KE_1 = 0
KE_2 = \frac{1}{2} \cdot m \cdot v^2
m = 24 \text{ kg}
\frac{1}{2} \cdot m \cdot v^2= 32000
v^2 = \frac{2 \cdot 32000}{m}
v = \sqrt{\frac{2 \cdot 32000}{24}} = 52 \frac{m}{s}
Work-Energy Principle
Energy is the