A mass of 0.5,kg moving with a speed of 1.5,m | vedclass

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(A) According to the law of conservation of energy,the kinetic energy of the mass is converted into the elastic potential energy of the spring at the

Vedclass · Accepted Answer

$0.15$

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(A) According to the law of conservation of energy,the kinetic energy of the mass is converted into the elastic potential energy of the spring at the point of maximum compression.Let $m = 0.5\,kg$ be the mass,$v = 1.5\,m/s$ be the velocity,and $k = 50\,N/m$ be the spring constant.The kinetic energy of the mass is $K.E. = \frac{1}{2}mv^2$.The potential energy stored in the spring at maximum compression $x$ is $P.E. = \frac{1}{2}kx^2$.Equating the two energies:$\frac{1}{2}mv^2 = \frac{1}{2}kx^2$Solving for $x$:$x^2 = \frac{mv^2}{k}$$x = \sqrt{\frac{mv^2}{k}} = v\sqrt{\frac{m}{k}}$Substituting the given values:$x = 1.5 	imes \sqrt{\frac{0.5}{50}}$$x = 1.5 	imes \sqrt{0.01}$$x = 1.5 	imes 0.1 = 0.15\,m$.Thus,the maximum compression is $0.15\,m$.

$A$ mass of $0.5\,kg$ moving with a speed of $1.5\,m/s$ on a horizontal smooth surface,collides with a nearly weightless spring of force constant $k = 50\,N/m$. The maximum compression of the spring would be ............. $m$.

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