Physics

Physics

Problem

A light horizontal spring is fixed to a wall on its left end. A small cart of mass mm is attached to the free end of the spring and can glide without friction along a straight track. The track’s origin x=0x=0 is defined at the spring’s equilibrium length. Three instants during an experiment are noted: • Instant 1: the cart is at x=+0.050 mx=+0.050\,\text{m} (to the right of equilibrium). • Instant 2: the cart is at x=0x=0. • Instant 3: the cart is at x=−0.050 mx=-0.050\,\text{m} (to the left of equilibrium). Answer the following questions about the restoring force exerted on the cart by the spring. (a) For each of the three instants, state the direction (left or right) of the restoring force acting on the cart. (b) Describe quantitatively how the magnitude of the restoring force depends on the cart’s displacement xx from equilibrium. Include the role of proportionality and any physical constant that characterizes the spring. (c) Without performing any numerical calculation, describe the overall shape, slope, and intercept of a graph of restoring force FspringF_{\text{spring}} (vertical axis) versus displacement xx (horizontal axis). Explain how the graph reflects both the magnitude–displacement relationship and the force’s direction.
Intuitive explanation

The restoring force in a spring system depends on displacement from equilibrium, not on position alone—what matters is how far the cart is stretched or compressed. At instants 1 and 3, the cart is equally displaced from equilibrium (just in opposite directions), so by Hooke's law (F=−kxF = -kx), the magnitudes of the restoring forces are identical, though their directions reverse. At instant 2, the cart passes through equilibrium where the spring exerts zero force, even though the cart is moving fastest. The key insight is separating the force (which depends only on current position) from the motion (which depends on velocity and energy).

Intuitive explanation

The restoring force in a spring system depends on displacement from equilibrium, not on position alone—what matters is how far the cart is stretched or compressed. At instants 1 and 3, the cart is equally displaced from equilibrium (just in opposite directions), so by Hooke's law (F=−kxF = -kx), the magnitudes of the restoring forces are identical, though their directions reverse. At instant 2, the cart passes through equilibrium where the spring exerts zero force, even though the cart is moving fastest. The key insight is separating the force (which depends only on current position) from the motion (which depends on velocity and energy).

Browse the problem library