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3 years ago

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A 2.5 kg block is launched along the ground by a spring with a spring constant of 56 N/m. The spring is initially compressed 0.7

5 m.
Disregarding friction, how fast will the block move after the spring is released all the way and the block slides away from it?

3.5 m/s
4.1 m/s
13 m/s
16 m/s

2 answers:

Norma-Jean [14]3 years ago

7 0

vf ^2 = kx^2/m = 56(0.75)^2 / 2.5 = 12.6

Therefore, v= 3.5 m/s.

olga55 [171]3 years ago

4 0

Answer: The correct option is 3.5m/s.

Explanation: In the question, conservation of energy is taking place which means that the energy is getting transferred from one form to another form.

Here, elastic potential energy of the spring is getting converted to the kinetic energy of the block.

Mathematically,

$\frac{1}{2}kx^2=\frac{1}{2}mv^2$

where, k = spring constant = 56N/m

x = compression of extension of spring = 0.75m

m = Mass of the block = 2.5kg

v = velocity of the block = ? m/s

Putting values in above equation, we get:

$\frac{1}{2}\times 56\times (0.75)^2=\frac{1}{2}\times 2.5\times v^2\\\\v=3.54m/s$

Hence, the block will move at a speed of 3.5m/s

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