A block of mass 16 kg is moving on a friction less horizontal surface with velocity 4 m/s and comes to rest after pressing a spring. If the force constant of the spring is 100 N/m then the compression in the spring will be?

3.2 m

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1.6 m

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0.6 m

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6.1 m

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Solution

The correct option is B $1.6 m$
The correct option is B.

Given,

$m a s s = 16 k g, v 3 e l o c i t y = 4 m / s, k = 10 N / m$

So,

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

$\frac{1}{2} \times 100 \times x^{2} = \frac{1}{2} \times 16 \times 4^{2}$

$50 x^{2} = 128$

$x^{2} = 2.56 = 1.6$

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Figure

A block of mass 2.0 kg is moving on a friction less horizontal surface with a velocity of 1.0 m/s below figure towards another block of equal mass kept at rest.The spring constant of the spring fixed at one end is 100 N/m.Find the maximum compression of the spring.

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A block of mass 16 kg is moving on a friction less horizontal surface with velocity 4 m/s and comes to rest after pressing a spring. If the force constant of the spring is 100 N/m then the compression in the spring will be?

The correct option is B 1.6mThe correct option is B.Given,mass=16kg,v3elocity=4m/s,k=10N/mSo,12kx2=12mv212×100×x2=12×16×4250x2=128x2=2.56=1.6

The correct option is B $1.6 m$
The correct option is B.

Given,

$m a s s = 16 k g, v 3 e l o c i t y = 4 m / s, k = 10 N / m$

So,

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

$\frac{1}{2} \times 100 \times x^{2} = \frac{1}{2} \times 16 \times 4^{2}$

$50 x^{2} = 128$

$x^{2} = 2.56 = 1.6$