Physics

Problem

A small cart of mass

0.50\,\text{kg}

moves frictionlessly along a straight horizontal track. The potential energy

U

of the cart–track system as a function of the cart’s position

x

(measured in meters along the track) is described verbally below. • At

x=0\,\text{m}

U=6\,\text{J}

. Moving in the

+x

direction,

U

decreases smoothly, reaching a minimum value of

2\,\text{J}

x=2\,\text{m}

. • Beyond

x=2\,\text{m}

the potential rises, attaining

U=5\,\text{J}

x=5\,\text{m}

, then remains constant at

5\,\text{J}

for

x>5\,\text{m}

. Assume the cart has negligible size and can move only along the

x

-axis. (a) Identify all positions of mechanical equilibrium for the cart and state whether each is stable or unstable. Explain your reasoning using the qualitative shape of the potential energy function. (b) At

x=3.0\,\text{m}

the cart is momentarily at rest. Describe the subsequent motion of the cart. In your answer, indicate the direction of the net force at that instant and the range of

x

-values through which the cart will move before turning around. (c) Determine the sign (positive, negative, or zero) of the change in potential energy

\Delta U

as the cart moves from

x=0.50\,\text{m}

x=1.5\,\text{m}

. Justify your answer without performing numerical calculations.

Intuitive explanation

This problem is about energy conservation and motion in a potential energy landscape. The key insight is that the cart's total mechanical energy (kinetic plus potential) stays constant as it moves, so you can use the shape of the $U(x)$ curve to determine where the cart can go and how fast it moves at each position. Think of the potential curve as a "hill" the cart rolls on—regions where $U$ is low mean the cart has more kinetic energy and moves faster, while steep slopes tell you where forces are strongest.

Intuitive explanation

Browse the problem library