Challenge your understanding of spring mechanics, force interactions, and physical constraints with these carefully crafted questions. This quiz covers Hooke's Law, force equilibrium, types of constraints, and practical applications relevant to mechanical and engineering systems.
A spring with a constant of 200 N/m stretches by 0.05 meters when a mass is hung from it. What is the magnitude of the force exerted by the mass on the spring?
Explanation: Using Hooke's Law, the force F equals the spring constant (k) times the extension (x), so F = 200 N/m × 0.05 m = 10 N. 1 N is too low, as it ignores the correct calculation. 4 N is also incorrect, resulting from a potential miscalculation. 20 N is double the correct value and does not align with the provided numbers. Only 10 N properly applies Hooke's Law.
If two people pull on opposite ends of a rope with equal forces of 120 N, what is the net force acting on the rope assuming it is stationary?
Explanation: Since the rope is stationary and both forces are equal and opposite, they cancel each other, leaving a net force of 0 N. 240 N is the sum but not the net force, which would move the rope. 120 N incorrectly considers only one force, and 60 N may result from averaging rather than summing correctly. Net force is zero here due to equilibrium.
Which type of mechanical constraint prevents an object from moving in both horizontal and vertical directions, but allows rotation?
Explanation: A pinned joint (or hinge) restricts horizontal and vertical translation but still permits rotation. A fixed joint prevents all movement, including rotation, making it incorrect. A roller support allows translation in one direction and is not as restrictive as described. Free body refers to an unconstrained object, which does not match the scenario.
If a spring is compressed by 0.2 meters and has a spring constant of 50 N/m, what is the elastic potential energy stored in the spring?
Explanation: The potential energy stored is (1/2)kx² = 0.5 × 50 × (0.2)² = 0.5 × 50 × 0.04 = 1.0 J. 2.0 J and 4.0 J result from forgetting to square the displacement or missing the 0.5 factor. 0.5 J ignores some of the multiplication required. The calculation leads only to 1.0 J as the correct answer.
What happens to the restoring force of a spring as it is stretched further from its equilibrium position, assuming the elastic limit is not exceeded?
Explanation: According to Hooke's Law, the restoring force increases proportionally to the displacement as long as the elastic limit is not reached. Decreasing exponentially is incorrect, as elasticity does not involve exponential decay in this context. A constant or zero force would not return the spring to equilibrium and ignores the principle behind Hooke’s Law. Thus, proportional increase is correct.