Question

A block having a mass of 0.80 kg is given an initial velocity 1.2 m/s to the right and collides with
spring of negligible mass and force constant k =50 N/m. Assuming the surface to be frictionless,
calculate the maximum compression of the spring after the collision. 1
a. 0.15 m b. 0.20m c.0.10 m . d. 0.25 m
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Answer 1

Maximum compression of the spring after the collision is 0.15 m.

Given :

Mass of block , m = 0.80 kg .

Initial velocity , u = 1.2 m/s .

Spring constant , k = 50 N/m .

Now, when maximum compression archive all kinetic energy will be converted in spring energy .

So,

        $K.E = E_{spring}\\\\\dfrac{mv^2}{2}=\dfrac{kx^2}{2}$

Putting all values in above equation.

We get ,

x = 0.15 m .

Hence , this is the required solution.

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Answer 2

The maximum compression of the spring after the collision is 0.15 m

Explanation:

Given data

Mass of the block (m) = 0.80 kg

Initial velocity (v) = 1.2 m/s

Spring constant (k) = 50 N/m

FInd the maximum compression of the spring (x) after compression

Potential energy of the spring = Kinetic energy of the block

Kinetic energy of the block = 0.5 × (mv)²

Kinetic energy of the block = 0.5 × (0.80 × 1.2)²

Kinetic energy of the block =0.5 × 0.9216

Kinetic energy of the block = 0.4608 ---------->(1)

Potential energy of the spring = 0.5 × k × x²

Potential energy of the spring = 0.5 × 50 × x²

Potential energy of the spring = 25 x² ---------> (2)

Equate (1) and (2)

25 x² = 0.4608

x² =  0.018432 m²

x =0.1357 = 0.15 m

Therefore the maximum compression of the spring after collision is 0.15 m  

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