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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 2 — The Loop and the Rough Ramp

Energy conservation through a frictionless loop, then dissipation on a rough incline.

A small toy block of mass \(m = 2\text{ kg}\) is pressed against a spring (spring constant \(k = 400\text{ N/m}\)), compressing it by \(x = 0.5\text{ m}\). The block is released from rest at Point A on a frictionless horizontal floor.

It passes through a frictionless, circular vertical loop of radius \(R = 1\text{ m}\) (Point B is at the very top of this loop).

After exiting the loop, the block slides up a ramp inclined at \(37^\circ\) above the horizontal. Unlike the rest of the track, the ramp has friction with a coefficient of kinetic friction \(\mu_k = 0.5\). The block travels up the ramp and comes to a momentary stop at Point C.

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

  1. As the block travels from the bottom of the loop to the top (Point B), the normal force acts on it. Calculate the work done by the normal force during the loop, and explain your reasoning.

  2. Calculate the speed of the block at the top of the loop (Point B).

  3. As the block slides up the inclined ramp, it loses energy to friction.

    • Calculate the magnitude of the friction force acting on the block on the ramp.
    • Write an equation for the work done by friction \(W_\text{fric}\) as a function of the distance \(d\) the block slides up the ramp.

  4. Use Conservation of Energy (accounting for non-conservative work) to calculate the total distance \(d\) the block slides up the ramp before coming to a stop at Point C.

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

- Topic: Energy conservation, including spring, kinetic, and gravitational potential energy.
- Difficulty Level: 4/5 (challenging with common pitfalls).
- Numbers: Optimized for no-calculator use.
- Previously Challenging Parts I Want to Study: "No net work" problems, and the non-intuitive effects of friction.

Prompt generated by Jens Mahlmann for Gemini 3.5 Flash.

Initial energy. The spring stores elastic potential energy, which converts entirely to kinetic energy on the frictionless floor:

\[E = \tfrac{1}{2}kx^2 = \tfrac{1}{2}(400)(0.5)^2 = 50\text{ J}\]

Since the floor is frictionless and horizontal, the block has \(50\text{ J}\) of kinetic energy entering the loop.

Part 1 — Work done by the normal force in the loop

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

The normal force is always directed toward the center of the loop, which is perpendicular to the block's velocity (and displacement) at every instant. Since \(W = \vec{F}\cdot\vec{d}\) and the angle is always \(90^\circ\), the normal force does no work.

Part 2 — Speed at the top of the loop (Point B)

Take the floor as \(h = 0\). The top of the loop is at height \(h = 2R = 2\text{ m}\). The loop is frictionless, so mechanical energy is conserved:

\[E = \tfrac{1}{2}mv_B^2 + mg(2R)\]
\[50 = \tfrac{1}{2}(2)v_B^2 + (2)(10)(2)\]
\[50 = v_B^2 + 40 \quad\Rightarrow\quad v_B^2 = 10 \quad\Rightarrow\quad v_B \approx 3.16\text{ m/s}\]

Part 3 — Friction on the ramp

The block returns to floor level with its full \(50\text{ J}\) of kinetic energy (the loop is frictionless, so no energy is lost there).

Friction force:

\[f_k = \mu_k F_N = \mu_k \, mg\cos\theta = 0.5(2)(10)(0.8) = 8\text{ N}\]

Work done by friction over a distance \(d\) up the ramp (opposing motion):

\[W_\text{fric} = -f_k \, d = -8d \quad\text{(in joules, }d\text{ in meters)}\]

Part 4 — Distance up the ramp (Point C)

Energy conservation with non-conservative work, from the floor to Point C (where the block momentarily stops, so final KE \(= 0\)):

\[E_\text{initial} + W_\text{fric} = \Delta PE\]

The block rises a height \(h = d\sin\theta\) along the ramp:

\[50 - 8d = mg\,d\sin\theta = (2)(10)(0.6)d = 12d\]
\[50 = 12d + 8d = 20d \quad\Rightarrow\quad d = 2.5\text{ m}\]

The block slides \(2.5\text{ m}\) up the ramp before stopping at Point C.

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!