Try each one on paper first, then check a single step. That is worth far more than reading a finished solution.
Example 1 — work done by a force at an angle
A box is dragged 8 m along the floor by a 50 N force acting at 30° to the horizontal. Find the work done by that force.
- 1W=Fdcosθ, with θ the angle between force and displacementOnly the component along the motion does work.
- 2=50(8)cos30°
- 3=400(0.8660)
- 4=346 JLess than 400 J, because part of the force lifts rather than pulls.
Example 2 — energy with friction present
A 60 kg child slides down a slide with a vertical drop of 5 m, starting from rest. Friction does 800 J of work. Find the speed at the bottom.
- 1PE lost =mgh=60(9.81)(5)=2943 JTake PE as zero at the bottom.
- 2Energy equation: PE lost = KE gained + work against frictionFriction removes energy, so it is subtracted from what is available.
- 321(60)v2=2943−800=2143 J
- 430v2=2143⇒v2=71.4
- 5v=8.45 m s−1Less than the frictionless 9.90 m s−1 ✓
Example 3 — power and acceleration
A car of mass 1200 kg has an engine working at 24 kW. At 20 m s−1 against a resistance of 500 N, find the acceleration and the maximum speed.
- 1P=Fv, so F=vP=2024000=1200 NThe driving force at that instant.
- 2Resultant =1200−500=700 NResistance opposes the motion.
- 3a=1200700=0.58 m s−2
- 4At maximum speed a=0, so the driving force equals the resistance: F=500 NNo resultant force means no acceleration.
- 5vmax=FP=50024000=48 m s−1As v rises at constant power, F falls.