Ablock of mass 0.25 kg is placed on a verti-cal spring of constant 3229 n/m and pusheddownward, compressing the spring 0.04 m. the block is released from this point. soonafter, it leaves the spring and continues totravel upward. the acceleration of gravity is 9.8 m/s2.what height above the point of release willthe block reach if air resistance is negligible? answer in units of m.

The height reached by the block is 1.06 m.

Explanation:

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According to the conservation of energy, all the elastic potential energy given by the compression of the string will be converted into gravitational potential energy when the block reaches its maximum height. At that point the velocity of the block is zero, then, the kinetic energy will be zero.

Before releasing the block, its mechanical energy (E) is equal to the elastic potential energy (EPE), plus the gravitational potential energy (PE) plus the kinetic energy (KE):

Initally:

E = EPE + PE + KE

E = EPE + 0 + 0

E = EPE

The elastic potential energy is calculated as follows:

EPE = 1/2 · k · x²

Where:

k = spring constant.

x = compression of the spring.

EPE = 1/2 · 3229 N/m · (0.04 m)²

EPE = 2.6 J

Then:

E = 2.6 J

The mechanical energy is constant because there are no dissipative forces applied on the block (like air resistance). Then, at the maximum height of the block, the mechanical energy will also be 2.6 J:

At the highest height:

E = EPE + PE + KE

E = 0 + PE + 0

2.6 J = PE

The potential energy is calculated as follows:

PE = m · g · h

Where:

m = mass of the block.

g = acceleration of gravity.

h = height.

Solving the equation for height:

PE /(m · g) = h

2.6 J /(0.25 kg · 9.8 m/s²) = h

h = 1.06 m

The height reached by the block is 1.06 m.

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