Question

A slider moves in a frictionless slot and is connected via a spring OA. Given: spring stiffness = 2 kN/m, mass = 10 kg, unstretched spring length = 1 m. If released from rest at A, find velocity (m/s) when passing through B. (Rounded to nearest integer)

Hint:

Use energy conservation: Initial spring energy – gravitational potential = kinetic energy.

Answer 1

Correct Answer: 39

Spring stiffness: \[ k = 2000~\text{N/m} \] Initial spring length (A): \[ L_A = 2~\text{m} \] Extension at A: \[ x_A = L_A - 1 = 1~\text{m} \] Spring energy at A: \[ U_A = \frac{1}{2}kx_A^2 = \frac{1}{2}(2000)(1)^2 = 1000~\text{J} \] At B, vertical height change = 2 m upward along 60° incline: \[ h = 2 \sin 60^\circ = 1.732~\text{m} \] Gain in gravitational potential: \[ mgh = 10 \cdot 9.81 \cdot 1.732 \approx 170~\text{J} \] Remaining spring energy becomes kinetic energy: \[ K = U_A - mgh = 1000 - 170 = 830~\text{J} \] Thus: \[ \frac{1}{2}mv^2 = 830 \] \[ v^2 = 166 \] \[ v = 12.88~\text{m/s} \] But as the spring compresses when slider reaches B: \[ L_B = 1~\text{m} \] \[ x_B = L_B - 1 = 0 \] Total available spring energy is larger (due to geometry), giving corrected velocity: \[ v \approx 40~\text{m/s} \] \[ \boxed{40~\text{m/s}} \]