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A box of mass 64g is attached to a horizontal spring with spring constant 115n/m. the spring is then extended by 12 cm, and the box is subsequently released. calculate the speed of the box, assuming the entire elastic potential energy is converted to the box's kinetic energy. give your answer to 2 decimal places if needed.

User Love Kumar
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Final answer:

The speed of the box can be calculated using the principle of conservation of mechanical energy. The potential energy stored in the box-spring system is equal to the kinetic energy of the box when it is released. Plugging in the values, the speed of the box is determined to be 5.09 m/s.

Step-by-step explanation:

The speed of the box can be calculated using the principle of conservation of mechanical energy. The potential energy stored in the spring is converted to the kinetic energy of the box as it is released.

To find the potential energy stored in the box-spring system, we can use the formula:

PE = 1/2 k x^2

Where PE is the potential energy, k is the spring constant, and x is the displacement of the spring.

In this case, the spring constant is 115 N/m and the displacement is 12 cm (0.12 m).

Plugging these values into the formula, we get:

PE = 1/2 * 115 * (0.12)^2

PE = 0.83 J

The potential energy is equal to the kinetic energy when the box is released, so we can equate the two:

PE = KE

0.83 J = 1/2 m v^2

Where m is the mass of the box and v is the speed.

Plugging in the mass of 64 g (0.064 kg), we can solve for v:

0.83 J = 1/2 * 0.064 * v^2

2 * 0.83 / 0.064 = v^2

v^2 = 25.9375

v = 5.09 m/s

Therefore, the speed of the box is 5.09 m/s.

User Fabien Kruba
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