Question

The spring has a stiffness of 1600 N/m and a relaxed length of 0.15 m . You pull the mass until the stretch of the spring is 0.08 m and release it from rest. The block begins to oscillate with negligible friction.

What is the block's speed when the stretch of the spring is 0.06 m?

Answer 1

Answer:

0.946 m/s

Explanation:

Given,

Spring stiffness, k = 1600 N/m

relaxed length, l = 0.15 m

spring is stretched to, A = 0.08 m

Speed of block at, x = 0.06 m

Using mass of the block be 5 Kg

Using Conservation of energy

$\dfrac{1}{2}kA^2 = \dfrac{1}{2}kx^2 + \dfrac{1}{2}mv^2$

$m v^2 = k (A^2-x^2)$

$v= \sqrt{\dfrac{k(A^2-x^2)}{m}}$

$v= \sqrt{\dfrac{1600\times (0.08^2-0.06^2)}{5}}$

$v = 0.946 m/s$

Hence, the speed of the block is equal to 0.946 m/s

Answer 2

The block's speed when the stretch of the spring is 0.06 m is mathematically given as

v = 0.946 m/s

What is the block's speed when the stretch of the spring is 0.06 m?

Question Parameter(s):

The spring has a stiffness of 1600 N/m and a relaxed length of 0.15 m .

The spring is 0.08 m and release it from rest.

Generally, the equation for the Conservation of energy  is mathematically given as

$0,5kA^2 =0.5kx^2 +0.5mv^2$

Therefore

m v^2 = k (A^2-x^2)

$v= \sqrt{\frac{k(A^2-x^2)}{m}}$

Hence

$v= \sqrt{\frac{1600* (0.08^2-0.06^2)}{5}}$

v = 0.946 m/s

In conclusion, the speed is

v = 0.946 m/s

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