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Midterm II Practice

This collection features solved problems covering work and energy, linear momentum and its conservation, and rotational dynamics—including rotational motion, moments of inertia, and angular momentum.

Problem 1 — The Diagonal Push

Resolving a horizontal force on an incline — and why the normal force (and thus friction) grows when you don't push parallel to the ramp.

A mover is pushing a \(10\text{ kg}\) crate up a ramp inclined at \(37^\circ\) above the horizontal. Instead of pushing parallel to the ramp, the mover exerts a constant horizontal force of \(100\text{ N}\) (parallel to the ground).

The crate slides a distance of \(5\text{ m}\) up the ramp. The coefficient of kinetic friction between the crate and the ramp is \(\mu_k = 0.1\).

(Use \(g = 10\text{ m/s}^2\), \(\sin 37^\circ = 0.6\), and \(\cos 37^\circ = 0.8\).)

  1. Find the component of the pushing force \(F\) that is parallel to the ramp, and the component that is perpendicular to the ramp.

  2. Calculate the magnitude of the Normal Force \(F_N\) acting on the crate. (Hint: it is not just \(mg\cos\theta\).)

  3. Calculate the work done on the crate by each of the four forces (Push, Gravity, Friction, and Normal Force).

  4. Use the Work–Energy Principle to find the net work done on the crate, and determine its final speed if it started from rest.

I need an exercise for exam practice for a General Physics I lecture at the college level (algebra-based).

- Topic: Work-energy principle, net work, and different forces (including vector projections).
- Difficulty Level: 3/5 (straightforward with minor pitfalls).
- Numbers: Optimized for no-calculator use.
- Previously Challenging Parts I Want to Study: Trigonometry in vector projections, and forces canceling out non-intuitively.

Prompt generated by Jens Mahlmann for Gemini 3.5 Flash.

Setup. Take the \(x\)-axis pointing up the ramp and \(y\) perpendicular to the ramp. The horizontal push \(F = 100\text{ N}\) must be resolved into these tilted axes.

Part 1 — Components of the push

A horizontal force makes angle \(\theta\) with the ramp surface:

\[F_\parallel = F\cos\theta = 100(0.8) = 80\text{ N (up the ramp)}\]
\[F_\perp = F\sin\theta = 100(0.6) = 60\text{ N (into the ramp)}\]

Part 2 — Normal Force

Perpendicular to the ramp, the crate is in equilibrium. Both gravity and the push have components pressing into the surface:

\[F_N = mg\cos\theta + F\sin\theta = 10(10)(0.8) + 100(0.6)\]
\[F_N = 80 + 60 = 140\text{ N}\]

This is why it's not just \(mg\cos\theta = 80\text{ N}\) — the horizontal push adds to the normal force.

Part 3 — Work done by each force (\(d = 5\text{ m}\) up the ramp)

Push: only the parallel component does work.

\[W_\text{push} = F_\parallel \, d = 80(5) = +400\text{ J}\]

Gravity: component along ramp is \(mg\sin\theta\), opposing motion.

\[W_\text{grav} = -mg\sin\theta \, d = -10(10)(0.6)(5) = -300\text{ J}\]

Friction: \(f_k = \mu_k F_N = 0.1(140) = 14\text{ N}\), opposing motion.

\[W_\text{fric} = -f_k \, d = -14(5) = -70\text{ J}\]

Normal Force: perpendicular to displacement.

\[W_\text{normal} = 0\text{ J}\]

Part 4 — Net work and final speed

\[W_\text{net} = 400 - 300 - 70 + 0 = +30\text{ J}\]

By the Work–Energy Principle, \(W_\text{net} = \tfrac{1}{2}mv^2 - 0\):

\[30 = \tfrac{1}{2}(10)v^2 \quad\Rightarrow\quad v^2 = 6 \quad\Rightarrow\quad v \approx 2.45\text{ m/s}\]

Feedback & Contributions

Have feedback, want to share your experience, or found a fun variation of this exercise? Email jens.f.mahlmann@dartmouth.edu — we'd love to hear from you!