a 60 kg bicyclist going 2 m/s increased his work output by 1,800 j. what was his final velocity? m/s

Understanding the Given Data

We’re given: Mass (m): 60 kg Initial velocity (v₁): 2 m/s Work done (ΔW): 1800 J Final velocity (v₂): Unknown

The keyword question — {{keyword}} — is ultimately a kinetic energy problem. Work done by the bicyclist gets converted into kinetic energy (assuming negligible losses to friction or heat).

Applying the WorkEnergy Principle

The workenergy principle says:

Work = Change in Kinetic Energy

Mathematically:

ΔKE = ½mv₂² – ½mv₁² = Work

We’ll use this to isolate v₂, the final velocity.

StepbyStep

Let’s plug in the numbers:

ΔKE = 1800 J m = 60 kg v₁ = 2 m/s

So:

1800 = ½ × 60 × v₂² – ½ × 60 × (2)²

1800 = 30v₂² – 30×4 1800 = 30v₂² – 120 Now add 120 to both sides:

1920 = 30v₂² Divide both sides by 30:

v₂² = 64 v₂ = √64 v₂ = 8 m/s

Final Answer for {{keyword}}

So the final velocity is 8 m/s. That extra 1,800 joules of work pushed the cyclist’s speed from 2 m/s up to 8 m/s — a solid gain driven entirely by the conversion of work into kinetic energy.

Why This Makes Sense

Work is energy. In this case, since the cyclist put out exactly 1,800 J more with no energy loss assumed, all of that energy became speed. Mass didn’t change. No hills or wind involved. Just raw acceleration via energy input.

Revisiting {{keyword}} in Physics Problems

A question like {{keyword}} isn’t about memorizing formulas. It’s about recognizing that energy input changes kinetic energy, which affects speed. If you understand that, you’re set for similar problems — with higher velocities, different masses, or varying amounts of work.

Quick Recap

Start from the workenergy principle Use the kinetic energy equation Plug in mass and known velocities Solve for the unknown speed

Once you’ve done that, you’ll see that something as abstract as “1,800 J” translates into a very literal push forward: nearly quadrupling this cyclist’s speed.

Final result: 8 m/s — that’s the cyclist’s new velocity after the work boost.