A spring with a spring constant of 1730 N/m is compressed 0.136 m from its equilibrium position with an object having a mass of

Question

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A spring with a spring constant of 1730 N/m is compressed 0.136 m from its equilibrium position with an object having a mass of

1.72 kg. The spring is released and the object moves along a frictionless surface when it reaches a small embankment. If the speed of the object is 2.45 m/s at location A, what is the embankment height h

Physics

1 answer:

Elan Coil [88] · Answer 1 · 2022-05-06T22:24:43+03:00

Given :

A spring with a spring constant of 1730 N/m is compressed 0.136 m from its equilibrium position with an object having a mass of 1.72 kg.

To Find :

The embankment in the height.

Solution :

Since no external force is acting in the system, therefore total energy will be conserved.

Initial kinetic energy of the object = Energy stored in spring

$K.E _i = \dfrac{kx^2}{2}\\\\K.E_i = \dfrac{1730\times 0.136^2}{2}\\\\K.E_i = 16\ J$

Also, initial potential energy is 0.

Now,

$K.E_i + P.E_i = K.E_f + P.E_f\\\\16 + 0 = \dfrac{1.72\times 2.45^2}{2}+ mgh\\\\mgh =16 - 5.16\\\\h = \dfrac{16 - 5.16}{1.72 \times 9.8}\\\\h = 0.64\ m$

Therefore, the embankment height is 0.64 m.