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

$0.15$

Vedclass · Answer

(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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