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Vaselesa [24]
3 years ago
11

A 0.025-kg block on a horizontal frictionless surface is attached to an ideal massless spring whose spring constant is The block

is pulled from its equilibrium position at x = 0.00 m to a displacement x = +0.080 m and is released from rest. The block then executes simple harmonic motion along the horizontal x-axis. When the displacement is what is the kinetic energy of the block?
Physics
1 answer:
DerKrebs [107]3 years ago
3 0

Complete Question

A 0.025-kg block on a horizontal frictionless surface is attached to an ideal massless spring whose spring constant is 150 N/m. The block is pulled from its equilibrium position at x = 0.00 m to a displacement x = +0.080 m and is released from rest. The block then executes simple harmonic motion along the horizontal x-axis. When the displacement is x = 0.024 m, what is the kinetic energy of the block?

Answer:

The kinetic energy is  KE = 0.4368\  J

Explanation:

From the question we are told that

   The mass of the block is m= 0.025\ kg

   The spring constant is k = 150 N/m

   The length of first  displacement  is x_1 = 0.80 \ m

     The length of first  displacement  is x_2 = 0.024 \ m

At the x_2 the kinetic energy is mathematically evaluated as

     KE  = \Delta E

Where \Delta E is the change in energy stored on the spring which is mathematically represented as

            \Delta E = \frac{1}{2} k (x_1 ^2 - x_2^2)

=>        KE = \frac{1}{2} k (x_1 ^2 - x_2^2)

Substituting value

          KE = \frac{1}{2} * 150 *  (0.08^2 - 0.024^2)

          KE = 0.4368\  J

   

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