A 2.+kg mass at the end of a spring with force-constant k = 875 N/m is pushed in

Question

A 2.+kg mass at the end of a spring with force-constant k = 875 N/m is pushed in, compressing the spring by 0.16 m, and then released. Find the elastic potential energy-, and use this to find the velocity right when

the mass leaves the (de-compressed) spring, (a) if the spring is horizontal [3.0 m/s], (b) if it's vertical; begin by calculating the gravitatibnal potential energy increase. (c) ln (b) find the miximum heightthe mass reaches above its initial stationary position [0.47 m. Since the mass is initially and finally stationary, kinetic energy needn't be considered here!l

Explanation / Answer

(a) Kinetic energy of block after leaving de compressed spring = Spring potential energy when completely compressed

0.5 mv^2 = 0.5 kx^2

0.5 (2)v^2 = 0.5 (875)(0.16)^2

v = 3.35 m/s [ with given data]

Answer will be 3.0 m/s if mass of block is 2.5 kg

0.5 (2.5)v^2 = 0.5 (875)(0.16)^2

v = 3.0 m/s [ with m = 2.5 kg]

(b) Kinetic energy gained by mass + Increase in potential energy of block= Spring potential energy when compressed

0.5 mv^2 + mgh = 0.5kx^2

0.5(2)v^2 + (2)(9.81)(0.16) = 0.5(875)(0.16)^2

v = 2.8 m/s [ with given data]

If mass = 2.5 kg then

0.5 mv^2 + mgh = 0.5kx^2

0.5(2.5)v^2 + (2.5)(9.81)(0.16) = 0.5(875)(0.16)^2

v = 2.4 m/s

(c) Final potential energy gained by mass at maximum height = Initial spring potential energy

mgh= 0.5 kx^2

where h is height above its stationary position

(2)(9.81)(h)= 0.5(875)(0.16)^2

19.62 h = 11.2

h = 0.57 m [ with given data]

If mass = 2.5 kg then

(2.5)(9.81)(h)= 0.5(875)(0.16)^2

24.5 h = 11.2

h = 0.46 m

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