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Lilit [14]
3 years ago
9

A mass m is tied to an ideal spring with force constant k and rests on a frictionless surface. The mass moves along the x axis.

Assume that x=0 corresponds to the relaxed position of the spring. The mass is pulled out to a position xm and released. Derive an expression for the positions at which the kinetic energy of the mass is equal to the elastic potential energy of the spring.
Physics
1 answer:
7nadin3 [17]3 years ago
3 0

Answer:x=\frac{x_m}{\sqrt{2}}

Explanation:

Given

initially mass is stretched to x_m

Let k be the spring Constant of spring

Therefore Total Mechanical Energy is \frac{kx_m^2}{2}

Position at which kinetic Energy is equal to Elastic Potential Energy

K=\frac{mv^2}{2}

U=\frac{kx^2}{2}

it is given

k=U

thus 2U=\frac{kx_m^2}{2}

2\times \frac{kx^2}{2}=\frac{kx_m^2}{2}

2x^2=x_m^2

x=\frac{x_m}{\sqrt{2}}

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Answer:

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Using the equation of motion

v^2 = u^2+2as \\ \\ v^2 = 2as \\ \\ v = \sqrt{2*a*s} \\ \\ v= \sqrt{2*4.1713*7.56} \\ \\ v = 7.942 \ m/s

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