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A ball of mass 2.00-kg is dropped from a height of 1.5 m (from the ground) onto a massless spring (the spring has an equilibrium length of 0.50 m). The ball compresses the spring by an amount of 0.20 m by the time it comes to a stop. Calculate the spring constant of the spring.

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Answer:

653.3 N/m

Step-by-step explanation:

From the question,

Potential energy of the falling ball = elastic energy stored in the spring.

mgh = 1/2ke²........................ Equation 1

Where m = mass of the ball, h = height from which the ball was dropped, g = acceleration due to gravity, k = spring constant, e = extension.

make k the subject of the equation

k = 2mgh/e²........................ Equation 2

Given: m = 2 kg, h = 1.5 m, g = 9.8 m/s², e = 0.5-0.2 = 0.3 m

Substitute these values into equation 2

k = 2(2)(9.8)(1.5)/0.3²

k = 653.3 N/m

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