Question

A 2.3 kg block is dropped from rest from a height of 4.6 m above the top of the spring. When the block is momentarily at rest, the spring is compressed by 25.0 cm. What is the speed of the block when the compression of the spring is 15.0 cm?

Answer 1

7.6 m/s

Step-by-step explanation:

m = 2.3 kg, h = 4.6 m, x = 25 cm = 0.25 m

Use the conservation of energy

Potential energy of the block = Elastic potential energy of the spring

m g h = 1/2 k x^2

Where, k be the spring constant

2.3 x 9.8 x 4.6 = 0.5 x k x (0.25)^2

k = 3317.88 N/m

Now, let v be the velocity of the block, when the compression is 15 cm.

Again use the conservation of energy

Potential energy of the block = Kinetic energy of block + Elastic potential

energy of the spring

m g h = 1/2 m v^2 + 1/2 k x^2

2.3 x 9.8 x 4.6 = 0.5 x 2.3 x v^2 + 0.5 x 3317.88 x (0.15)^2

103.684 = 1.15 v^2 + 37.33

v = 7.6 m/s