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chubhunter [2.5K]
3 years ago
14

2. A block of mass 1.2 kg lies on a frictionless surface. A man slides the block

Physics
1 answer:
Alex17521 [72]3 years ago
3 0

Answer:

The spring constant will be "1333.33 N/m".

Explanation:

The given values are:

Mass,

m = 1.2 kg

Displacement compression,

x = 0.15 m

Block's velocity,

v_i=5 \ m/s

As we know,

⇒  E_i=E_f

or,

⇒  K_i+v_i=K_f+v_f

⇒  \frac{1}{2}mv_i^2+0=0+ \frac{1}{2}Kx^2

So,

⇒  K=\frac{mv_i^2}{x_2}

On substituting the values, we get

⇒       =\frac{1.2\times 5\times 5}{0.15\times 0.15}

⇒       =\frac{30}{0.0225}

⇒       =1333.33 \ N/m

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