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:
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:
[tex]K=\frac{2mgh}{x^{2} } =\frac{2*2.1*9.8*5.5}{0.25^{2} } =3622.08N/m[/tex]
The total energy is equal to:
[tex]E_{total} =mgh=2.1*9.8*5.5=113.19J[/tex]
The speed of the block is:
[tex]v=\sqrt{\frac{2E_{total}-kx_{1}^{2} }{m} } =\sqrt{\frac{(2*113.19)-(3622.08*0.15^{2} )}{2.1} } =8.31m/s[/tex]
