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M016 · Newtonian contact mechanics

Accelerating Wedge & Sliding Block

Can a frictionless normal force transfer energy? Yes. In the ground frame, a freely recoiling wedge gains the energy that the normal force removes from the block. A horizontal drive can reverse this transfer. In the wedge frame, the normal force does no work on the relative slide; the energy balance includes inertial-force work.

Release or horizontally drive a frictionless wedge. Compare ground and wedge observers at the same instant, with consistent force arrows, inertial forces, kinetic energies and work.

Interactive modelAccelerating Wedge & Sliding Block
Recorded time0.228 s0.228\,\mathrm{s}
Distance down the face1.6 m1.6\,\mathrm{m}
Velocity relative to the face1.79 m s−11.79\,\mathrm{m\,s^{-1}}
Acceleration along the face7.85 m s−27.85\,\mathrm{m\,s^{-2}}
Normal force on block6.8 N6.8\,\mathrm{N}
Block speed in selected frame1.18 m s−11.18\,\mathrm{m\,s^{-1}}
Block horizontal acceleration in selected frame3.4 m s−23.4\,\mathrm{m\,s^{-2}}
Horizontal inertial force on block0 N0\,\mathrm{N}
Inertial-force work on block0 J0\,\mathrm{J}
Wedge displacement−0.0883 m-0.0883\,\mathrm{m}
Wedge velocity−0.775 m s−1-0.775\,\mathrm{m\,s^{-1}}
Wedge acceleration in ground frame−3.4 m s−2-3.4\,\mathrm{m\,s^{-2}}
Block kinetic energy0.7 J0.7\,\mathrm{J}
Wedge kinetic energy0.3 J0.3\,\mathrm{J}
Normal work in selected frame−0.3 J-0.3\,\mathrm{J}
Horizontal external work0 J0\,\mathrm{J}
Total horizontal momentum0 kg m s−10\,\mathrm{kg\,m\,s^{-1}}
Horizontal impulse residual0 kg m s−10\,\mathrm{kg\,m\,s^{-1}}
Record end time0.814 s0.814\,\mathrm{s}
Energy-balance residual0 J0\,\mathrm{J}

Physics tutorial

The same contact transfers energy differently in different frames

BackgroundA freely moving wedge recoils as the block slides down. On the ground, the normal force can transfer energy between them; riding with the wedge changes the displacement used to calculate that work.

Why it mattersWatch one motion from the ground and from the moving base. OpenStax 5.7 and 9.3 provide the force and momentum principles; MIT Engineering Dynamics lecture 8 explains inertial forces in accelerating observers.

Start with the essentials

Focus question
Why does normal-force work change when the observer rides with the wedge?
One-sentence intuition
The normal is perpendicular to the slide relative to the face. Its ground-frame work can be nonzero because the face moves too; the accelerating frame needs inertial-force work in its energy balance.

Core mathematical model

Two coordinates describe the motion

xb=X+(s−ℓ/2)cos⁡θyb=(ℓ−s)sin⁡θ\begin{aligned}x_b&=X+(s-\ell/2)\cos\theta\\y_b&=(\ell-s)\sin\theta\end{aligned}

The wedge moves horizontally; the block coordinate increases downhill along its face. The drawn body has a fixed schematic normal offset, which changes no acceleration or energy difference.

Coupled equations of motion

(M+m)X¨+mcos⁡θ s¨=Hs¨+cos⁡θ X¨=gsin⁡θ\begin{aligned}(M+m)\ddot X+m\cos\theta\,\ddot s&=H\\\ddot s+\cos\theta\,\ddot X&=g\sin\theta\end{aligned}

The horizontal drive acts on the wedge. The floor and face are frictionless, and the assembly has no rotational degree of freedom.

Recoil and contact force

X¨=H−mgsin⁡θcos⁡θM+msin⁡2θN=m(gcos⁡θ+X¨sin⁡θ)\begin{aligned}\ddot X&=\frac{H-mg\sin\theta\cos\theta}{M+m\sin^2\theta}\\N&=m(g\cos\theta+\ddot X\sin\theta)\end{aligned}

A free wedge accelerates left. A suitable rightward drive can reverse the recoil or make the block slide uphill relative to the wedge. Contact remains compressive over the available ranges.

Momentum and energy in the ground frame

px=(M+m)X˙+ms˙cos⁡θp˙x=HΔ(Kb+Kw+U)=HΔX\begin{aligned}p_x&=(M+m)\dot X+m\dot s\cos\theta\\\dot p_x&=H\\\Delta(K_b+K_w+U)&=H\Delta X\end{aligned}

With no horizontal drive, total horizontal momentum and mechanical energy stay constant. Each kinetic energy uses the full ground-frame velocity.

Contact work on the two bodies

WN,b=Nsin⁡θ ΔXWN,w=−WN,b\begin{aligned}W_{N,b}&=N\sin\theta\,\Delta X\\W_{N,w}&=-W_{N,b}\end{aligned}

During free recoil, the block loses energy through contact and the wedge gains it. A drive can reverse the transfer; the two contact-work contributions still cancel.

Riding with the accelerating wedge

ab′=ab−X¨x^Fin,b=−mX¨x^Δ(Kb′+U)=Win,b\begin{aligned}\mathbf a_b^{\prime}&=\mathbf a_b-\ddot X\hat{\mathbf x}\\\mathbf F_{\mathrm{in},b}&=-m\ddot X\hat{\mathbf x}\\\Delta(K_b^{\prime}+U)&=W_{\mathrm{in},b}\end{aligned}

This observer translates without rotating. Normal-force work on the relative slide is zero; the inertial force accounts for the relative energy balance. Switching observers preserves the same physical trial and current time.

Common difficulties

The base motion changes the slide

Typical misconceptionThe fixed-incline acceleration applies to every wedge.

Better mental modelA moving base accelerates the face. Compare Free wedge with Heavy base; the recoil and relative slide both change.

Work depends on the observer

Typical misconceptionA normal force can never transfer energy.

Better mental modelLook at normal work in the ground frame, then switch to the wedge frame. The relative slide is perpendicular to the normal, while the ground displacement includes base motion.

Inertial forces belong to an accelerating description

Typical misconceptionThe violet arrow is another physical contact.

Better mental modelViolet dashed arrows appear in the wedge frame. They account for the observer acceleration; the teal contact forces are the same physical interaction in either view.

Run the experiment

  1. 01

    Release both bodies

    Watch the Free wedge preset.

    What to observe: The base recoils left as the block slides down. Ground-frame normal work on the block is negative.
  2. 02

    Ride with the wedge

    Pause, then choose Wedge frame.

    What to observe: The base stays still and the ground datum moves. Violet inertial forces appear, and normal work on the block becomes zero.
  3. 03

    Keep the same instant

    Switch between the two observers.

    What to observe: The trial and time are preserved. The selected-frame speed, work and kinetic energies change together.
  4. 04

    Drive the block uphill

    Choose Driven uphill and inspect the relative velocity.

    What to observe: The rightward acceleration of the support drives relative uphill motion. External work and impulse account for the ground-frame balance.
  5. 05

    Keep the block at one place on the face

    Choose Ride with the wedge.

    What to observe: Both bodies accelerate horizontally while their relative position stays fixed.
  6. 06

    Catch a relative turning point

    Choose Uphill release. Open Work, constraints and diagnostics and press Exact zero speed, then Step 1 ms.

    What to observe: The relative velocity reaches zero and changes sign. The block starts sliding down the face again.
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