Kinetic Energy
What is the formula for kinetic energy?
KE = ½mv²
What is the formula for work?
W = Fdcos0
work= (force)(distance)(cos theta)
What is the Work-Energy Theorem?
W_net = ΔKE = 1/2mv^2f - 1/2mv^2i
What does it mean if work is positive?
Energy is added to the system
What is the base unit breakdown of a Joule?
1 J = 1 kg·m²/s²
What two factors does kinetic energy depend on?
Mass and speed (velocity).
If an object moves perpendicular to a force, how much work is done?
0 Joules
A 3 kg ball speeds up from 0 to 10 m/s. How much net work was done on it?
½(3)(10²) = 150 J
What’s the sign of work done by friction on a sliding object?
Negative
Define power
The rate at which work is done or energy is transferred.
Is kinetic energy a scalar or vector quantity?
Scalar
A box is pushed with 15 N at a 60° angle to the floor for 4 meters. How much work is done by the force?
15 × 4 × cos(60°) = 30 J
If net work is negative, what happens to the object’s speed and kinetic energy?
Both decrease
A student lifts a 15 kg box upward. What’s the sign of the work done by gravity?
Negative – gravity opposes the upward motion.
A force-distance graph shows a constant 5 N force over 10 m. What does the area under the graph represent?
Work = 50 J
Calculate the KE of a 10 kg object moving at 3 m/s.
KE = ½(10)(3²) = 45 J
A 20 N force is applied to a crate at a 45° angle over 6 m. What is the work done?
20 × 6 × cos(45°) ≈ 84.9 J
A 2 kg object is moving at 10 m/s. A constant force of 20 N brings it to rest over a distance. How far did it travel before stopping?
100 = 20d → d = 5 m
An object moves horizontally. Which forces do no work on it and why?
Gravity and normal force – no displacement in their direction.
An object moves at constant speed despite being pushed. What does this say about net force and net work?
Both are zero
A 12 kg object slows from 18 m/s to 4 m/s. What is the change in kinetic energy?
ΔKE = ½(12)(4²) - ½(12)(18²) = 96 - 1944 = -1848 J
A box is pushed with 35 N across a floor at constant speed for 6 m. What is the total net work done on the box?
0 J
(constant speed ⇒ ΔKE = 0 ⇒ Wnet = 0)
A 6 kg cart slows from 8 m/s to 2 m/s. Use the Work-Energy Theorem to find net work.
ΔKE = (1/2)(6)(2²) - (1/2)(6)(8²)
ΔKE = (1/2)(6)(4 - 64)
ΔKE = 3(-60) = -180 J
A 10 N force is applied at 60° to the direction of motion over 4 m. What component of the force does work, and what is the work value?
Only the horizontal component (cos 60°): W = 10 × 4 × 0.5 = 20 J
A 6 kg object slows from 10 m/s to 2 m/s while sliding across a surface. How much energy was lost to friction?
ΔKE = ½(6)(4 - 100) = -288 J → 288 J lost to friction