Work, Energy and Power: Conceptual Doubts Cleared (2)

medium • 3 min read
Tags Work Energy Power

Question

A block of mass mm slides down a frictionless incline of angle θ\theta from height hh. Two students argue: Student A says the work done by gravity is mghmgh. Student B says it is mghcos⁡θmgh\cos\theta because we should use the displacement along the incline. Who is right and why?

Solution — Step by Step

Work done by a constant force is W=F⃗⋅s⃗=Fscos⁡ϕW = \vec{F}\cdot\vec{s} = Fs\cos\phi, where ϕ\phi is the angle between the force and the displacement. The force here is gravity (mgmg downward); the displacement is along the incline of length L=h/sin⁡θL = h/\sin\theta.

Gravity points straight down. The displacement vector along the incline makes angle (90∘−θ)(90^\circ - \theta) with the vertical (because the incline makes angle θ\theta with the horizontal). So ϕ=90∘−θ\phi = 90^\circ - \theta and cos⁡ϕ=sin⁡θ\cos\phi = \sin\theta.

Wgravity=mg⋅L⋅sin⁡θ=mg⋅hsin⁡θ⋅sin⁡θ=mghW_{\text{gravity}} = mg \cdot L \cdot \sin\theta = mg \cdot \frac{h}{\sin\theta} \cdot \sin\theta = mgh

So Student A is right. Work done by gravity is mghmgh, regardless of the angle of the incline.

Why This Works

This is the entire point of gravity being a conservative force — work depends only on the change in vertical height, not the path. Whether the block slides down a steep slope, a gentle slope, or falls straight down, gravity does the same work mghmgh as long as it descends by the same vertical drop hh.

Student B’s mistake is using cos⁡θ\cos\theta instead of finding the actual angle between the force and displacement vectors. The angle between vertical (gravity) and the incline-direction is 90∘−θ90^\circ - \theta, so the cosine of that is sin⁡θ\sin\theta — and Lsin⁡θ=hL\sin\theta = h exactly.

Whenever you compute work done by gravity, ignore the path. Just find the vertical drop and multiply by mgmg. This is why energy conservation is so powerful — it skips the geometry entirely.

Alternative Method

Use the work-energy theorem. At the bottom, the block has speed vv where 12mv2=mgh\tfrac{1}{2}mv^2 = mgh (since the incline is frictionless and normal force does zero work). The kinetic energy gain equals mghmgh, so the only force doing work — gravity — must have done exactly mghmgh.

Students confuse the angle in W=Fscos⁡ϕW = Fs\cos\phi with the angle of the incline. The ϕ\phi in the formula is always the angle between F⃗\vec{F} and s⃗\vec{s}, never just “the angle in the diagram.”

Final answer: Student A is correct. Work done by gravity is mghmgh.

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