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3 years ago

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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, t

he spring is compressed by 25.0 cm. What is the speed of the block when the compression of the spring is 15.0 cm

2 answers:

ella [17]3 years ago

8 0

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

azamat3 years ago

3 0

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:

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

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