- Home
- Standard 12
- Physics
2. Electric Potential and Capacitance
medium
A block of mass $m$ moving with speed $v$ compresses a spring through distance $x$ before its speed is halved. What is the value of spring constant ?
A
$\frac{{3m{v^2}}}{{4{x^2}}}$
B
$\frac{{m{v^2}}}{{4{x^2}}}$
C
$\frac{{m{v^2}}}{{2{x^2}}}$
D
$\frac{{2m{v^2}}}{{{x^2}}}$
Solution
The speed of the block is halved after coving distance $x$.
The Initial Kinetic Energy of the block is $=\frac{1}{2} m v^{2}$
The Final Kinetic Energy of the block is $=\frac{1}{2} m(v / 2)^{2}$
The Change in Kinetic Energy is given as $\Delta K E=-\frac{1}{2} m\left(\frac{3 v^{2}}{4}\right)$
So PE + Change in $\mathrm{KE}=0$
The PE stored in the spring is $P E=\frac{1}{2} k x^{2}=\frac{3 m v^{2}}{8}$
$k=\frac{6 m v^{2}}{8 x^{2}}=\frac{3 m v^{2}}{4 x^{2}}$
Standard 12
Physics
