We are given the following information:
When the ball collides with the horizontally mounted spring, its kinetic energy is completely converted into potential energy stored in the spring. The potential energy stored in the spring at maximum compression is given by: \[ E_{\text{spring}} = \frac{1}{2} k x^2 \] Where: - \( k \) is the spring constant, - \( x \) is the maximum compression of the spring (0.5 m). Since all the kinetic energy is transferred to the spring, we equate the kinetic energy to the potential energy stored in the spring: \[ 10^3 \, \text{J} = \frac{1}{2} k (0.5)^2 \] Solving for \( k \): \[ 10^3 = \frac{1}{2} k (0.25) \] \[ 10^3 = 0.125 k \] \[ k = \frac{10^3}{0.125} = 8 \times 10^3 \, \text{N/m} \]
Correct Answer: (C) \( 8 \times 10^3 \, \text{N/m} \)
A ball is projected in still air. With respect to the ball the streamlines appear as shown in the figure. If speed of air passing through the region 1 and 2 are \( v_1 \) and \( v_2 \), respectively and the respective pressures, \( P_1 \) and \( P_2 \), respectively, then
If the voltage across a bulb rated 220V – 60W drops by 1.5% of its rated value, the percentage drop in the rated value of the power is: