Work and energy flashcards

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Work done (by a force)
The force multiplied by the distance moved in the direction of the force. A scalar, unit J.
The joule
The work done when a force of 1 N moves its point of application 1 m in the direction of the force. 1 J = 1 N m = 1 kg m² s⁻².
Kinetic energy
The energy a body has because of its motion: Ek = ½mv².
Gravitational potential energy
The energy a body has because of its position in a gravitational field; near the Earth's surface it changes by mgΔh.
Principle of conservation of energy
Energy cannot be created or destroyed; it can only be transferred from one store to another, so the total energy of a closed system is constant.
Resistive force
A force that opposes the motion, such as friction or air resistance; it does negative work and transfers energy to the surroundings.
Braking distance
The distance a vehicle travels while the brakes are applied, until it stops. With a constant braking force F: Fd = ½mv².
Work done by a constant force along the displacement
ΔW = FΔs (Yes)
Work done by a force F at angle θ to the displacement
ΔW = FΔs cos θ (component of the force along the displacement × displacement). (No — learn it)
Equation for kinetic energy
Ek = ½mv² (Yes)
Change in gravitational potential energy near the Earth's surface
ΔEgrav = mgΔh, with Δh the vertical height change. (Yes)
Speed after a vertical drop Δh with no resistive forces
v² = u² + 2gΔh (from mgΔh = ½mv² − ½mu²; the mass cancels). (No — derive it)
Energy balance for a body moving down with a resistive force
Loss of GPE = gain in KE + work done against resistive forces: mgΔh = ΔEk + F_r × d. (No — derive it)
Vertical drop of a pendulum bob released at angle θ
h = L(1 − cos θ), where L is the string length. (No — derive it)
Area under a force–displacement graph
The work done by the force (constant or varying).
Work done by a varying force from a curved F–s graph
Count the squares under the curve and multiply by the work represented by one square.
Shape of a graph of kinetic energy against speed
A parabola through the origin (Ek ∝ v²).
Shape of a graph of braking distance against initial speed (constant braking force)
A parabola through the origin: d = mv²/2F, so d ∝ v².
Graph of GPE against height near the Earth's surface
A straight line through the origin with gradient mg.
Ek and Egrav against height for a body falling with no air resistance
Egrav rises linearly with height, Ek falls linearly; their sum is a horizontal line.
Why does a force at 90° to the motion do no work?
It has no component along the displacement, so the body moves no distance in the direction of the force.
Why is the work done by friction on a sliding body negative?
Friction acts opposite to the displacement (θ = 180°, cos θ = −1), so it removes energy from the body.
Why does a ramp make lifting easier without saving energy?
The force along the ramp is smaller, but it acts over a longer distance; the work done equals mgΔh either way.
Total work done on a body moving at constant velocity
Zero: the resultant force is zero and the kinetic energy does not change.
Why does a projectile's landing speed not depend on the launch angle (no air resistance)?
Kinetic energy is a scalar, and the loss of GPE depends only on the vertical drop, so v² = u² + 2gΔh for any angle.
Where does the kinetic energy go when a car brakes?
The braking force does work on the car; the energy is transferred to thermal energy in the brakes and then to the surroundings.
Why do a pendulum's swings get smaller?
Work is done against air resistance, transferring energy to the surroundings, so less energy is left for each swing.
A 50 N pull at 25° to the horizontal drags a box 8.0 m along level ground. Work done?
50 cos 25° × 8.0 = 360 J (363 J).
Kinetic energy of a 1500 kg car at 20 m s⁻¹?
½ × 1500 × 20² = 3.0 × 10⁵ J. (Yes)
A 55 kg student climbs stairs that rise 3.2 m. Gain in GPE?
55 × 9.81 × 3.2 = 1.7 × 10³ J (1730 J). (Yes)
A ball is dropped from rest through 1.25 m with no air resistance. Speed on landing?
v = √(2 × 9.81 × 1.25) = 5.0 m s⁻¹. (No — derive it)
A 40 kg child slides from rest down a 5.0 m high slide and reaches 8.0 m s⁻¹. Energy transferred to the surroundings?
40 × 9.81 × 5.0 − ½ × 40 × 8.0² = 1962 − 1280 = 680 J.
A 900 kg car at 15 m s⁻¹ brakes with a constant 4500 N force. Braking distance?
Fd = ½mv² → d = (½ × 900 × 15²) ÷ 4500 = 23 m (22.5 m).
F–s graph: force rises from 0 to 300 N over 0.20 m, then stays at 300 N for 0.30 m. Work done?
Area = ½ × 0.20 × 300 + 0.30 × 300 = 30 + 90 = 120 J.
Δh in mgΔh for a slope of length L at angle θ
Δh = L sin θ — the vertical height, never the length along the slope.
Do you need the mass to find a speed by energy?
Not with negligible resistive forces — it cancels. You do need it for energy transferred or a resistive force.
Speed doubles — what happens to Ek and braking distance?
Both multiply by four (Ek ∝ v²).
Units before substituting into ½mv²
Mass in kg, speed in m s⁻¹ (divide km h⁻¹ by 3.6). A unit slip in v is squared.
Distance used for work done against a resistive force
The distance moved along the path, not the vertical height.
Laying out a conservation of energy answer
Write the balance in words first (loss of GPE = gain in KE + energy to surroundings), then substitute.

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