Question

A 2.1 kg block is dropped from rest from a height of 5.5 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

Answer:

The speed of the block is 8.2 m/s

Explanation:

Given;

mass of block, m = 2.1 kg

height above the top of the spring, h = 5.5 m

First, we determine the spring constant based on the principle of conservation of potential energy

¹/₂Kx² = mg(h +x)

¹/₂K(0.25)² = 2.1 x 9.8(5.5 +0.25)

0.03125K = 118.335

K = 118.335 / 0.03125

K = 3786.72 N/m

Total energy stored in the block at rest is only potential energy given as:

E = U = mgh

U = 2.1 x 9.8 x 5.5 = 113.19 J

Work done in compressing the spring to 15.0 cm:

W = ¹/₂Kx² = ¹/₂ (3786.72)(0.15)² = 42.6 J

This is equal to elastic potential energy stored in the spring,

Then, kinetic energy of the spring is given as:

K.E = E - W

K.E = 113.19 J - 42.6 J

K.E = 70.59 J

To determine the speed of the block due to this energy:

KE =  ¹/₂mv²

70.59 =  ¹/₂ x 2.1 x v²

70.59 = 1.05v²

v² = 70.59 / 1.05

v² = 67.229

v = √67.229

v = 8.2 m/s

Answer 2

Answer:

The speed of the block when the compression of the spring is 15 cm is 8.31 m/s

Explanation:

Given:

m = 2.1 kg

h = 5.5 m

x = 25 cm = 0.25 m

x₁ = 15 cm = 0.15 m

The energy is equal to:

$K=\frac{2mgh}{x^{2} } =\frac{2*2.1*9.8*5.5}{0.25^{2} } =3622.08N/m$

The total energy is equal to:

$E_{total} =mgh=2.1*9.8*5.5=113.19J$

The speed of the block is:

$v=\sqrt{\frac{2E_{total}-kx_{1}^{2} }{m} } =\sqrt{\frac{(2*113.19)-(3622.08*0.15^{2} )}{2.1} } =8.31m/s$