Question

A 2 kg block is attached to a horizontal ideal spring with a spring constant of 200N/m.when the spring has it`s equilibrium length the block is given a speed of 5m/s.what is the maximum elongation of the spring ?

Answer 1

Explanation:

pressure = 200 N/m^2

mass= 2 kg

speed = 5 m/s

as we know ,

2gh=v^2-u^2

2×9.8×h=5^2-0

2×9.8h=25

h =25 ×10/2×9.8

=125/98

=1.27meter

Answer 2

Answer:

0.5 m

Explanation:

at least elongation block has only kinetic energy and at most elongation it has only potential energy...

Therefore,

$\frac{1}{2} m {v}^{2} = \frac{1}{2} k {x}^{2}$

$\frac{1}{2} \times 2 \times {5 }^{2} = \frac{1}{2} \times 200 \times {x}^{2}$

$2 \times 25 = 200 \times {x}^{2}$

$\frac{50}{200} = {x}^{2}$

${x}^{2} = \frac{1}{4}$

$x = \sqrt{ \frac{1}{4} }$

$x = \frac{1}{2}$

$x = 0.5$

mark brainiest and follow me