Question

The figure shows an 8.5 kg stone at rest on a spring. The spring Is compressed 11 cm by the stone. What is the spring constant? The stone is pushed down an additional 33 cm and released. What is the elastic potential energy of the compressed spring just before that release? What is the change in the gravitational potential energy of the stone-Earth system when the stone moves from the release point to its maximum height? What is that maximum height, measured from the release point?

Answer 1

If y is the compression in a spring due to a force F, then F = – k y.

Also, change in potential energy of a spring when it is compressed by a distance y, U = (1/2) k y² .

1) The force acting on the spring: F = mg (weight of the stone).

    Then, mg = k y (numerically)     ==> k = mg / y

Here, m = 8.5 kg, y = 11 cm = 0.11 m, g = 9.8 m/s².

Therefore, k = 8.5 × 9.8 / 0.11 = 757.27 N/m = 757.3 N/m.

2) The spring is compressed by an ADDITIONAL 33 cm. The term ADDITIONAL is the key. This means that in addition to the 11 cm compressed earlier, an additional 33 cm compression is also effected. This makes total compression to be 11 + 33 = 44 cm.

That is, in this case,    y = 44 cm = 0.44 m. k = 757.3 N/m (just calculated).

Change in eleastic potential energy of the spring for a total compression of y;   U = (1/2) k y²

i.e.,   U = (1/2) × 757.3 × 0.44² = 73.3 J

3) On release, the energy stored in the spring is convered to the gravitational potential energy of the stone. Assuming no dissipation in energy, change in gravitational potential energy of the earth-stone system:

U = 73.3 J

4) The stone will reach upto a height where its gravitational potential energy (mgh) is equivalent to the PE given to it.

i.e., m g h = 73.3 J

or, 8.5 × 9.8 × h = 73.3 J   ==> h = 73.3 / (8.5 × 9.8) = 0.88 m = 88 cm.

Hope it is clear...