Question

A block of 9 kg is moving with velocity of 4 m/s on frictionless surface.
The block comes into rest by compressing the spring. If spring constant is
4x10' N/kg then find the compression in the spring.
​

Answer 1

HEYA GM ❤

★ BY CONSERVATION OF MECHANICAL ENERGY

∆ 1/2 mv² + mgh = 1/2 m2v²2 + mgh + 1/2Kx²

∆ 1/2mv² + 0 = 0 + 0 +1/2 Kx²

∆ 1/2mv² = 1/2Kx²

∆ 1/2 × 9 × 4² = 1/2 Kx²

∆ 9 × 8 = 1/2 × 4 × 10⁴ × x²

∆ 72 = 2 × 10² × x²

∆ 72 / 200 = x²

★ x² = 0.36 cm

★★ x = 0.6cm

HENCE COMPRESSION IS 0.6 CM IN SPRING....

THANK YOU ^_^✌

Answer 2

Given :

Mass of block , m = 9 kg .

Velocity of block , v = 4 m/s .

Spring constant , $k=4\times 10^4\ N/m$ .

To Find :

The compression in the spring.

Solution :

We know , in spring mass system total energy is conserved .

Therefore , Initial energy = Final energy

$\dfrac{mv_i^2}{2}+\dfrac{kx_i^2}{2}=\dfrac{mv_f^2}{2}+\dfrac{kx_f^2}{2}\\\\9\times 4^2+0=0+4\times 10^4 \times x_f^2\\\\x_f=\sqrt{\dfrac{9\times 16}{4\times 10^4}}\\\\x_f=0.06\ m$

Therefore , the compression in the spring is 0.06 m and 6 cm .

Learn More :

Conservation of energy

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