A block of mass 10 kg is attached to a spring of stiffness 360 N/m. A velocity of 6 m/s is given to the mass when the spring is in unstretched condition. The block will come to rest after moving a distance of

Question

Question

This question was previously asked in

UPPSC AE Mechanical 2022 Official Paper I (Held on 29 May 2022)

Answer 1

Concept

As the block moves the kinetic energy of the block decreases and the potential energy of the spring increases
At the extreme point total kinetic energy of the block is converted into the potential energy of the spring \(\frac{1}{2} mv^{2} \) = \(\frac{1}{2}kx^{2}\)

Calculation:

Given:

F1 Madhuri Engineering 08.08.2022 D14

K = 360N/m, m=10 Kg, V = 6m/s

\(\frac{1}{2} mv^{2}\) = \(\frac{1}{2}kx^{2}\)

\( 10\times6^{2}\) = \( 360\times x^{2}\)

x = 1m