The maximum compression produced in a spring of force constant $k$ , when a block of weight $W$ moving with velocity $v$ on a frictionless horizontal surface compresses it, shall be:

v \sqrt{\frac{W}{k}}

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v \sqrt{\frac{k}{W}}

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v \sqrt{\frac{W}{k . g}}

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v \sqrt{\frac{k . g}{W}}

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Solution

The correct option is C $v \sqrt{\frac{W}{k . g}}$
From conservation of energy,
KE lost by the block $= P E$ gained by the spring
so $\frac{1}{2} m v^{2} = \frac{1}{2} k d^{2}$
or $d = \sqrt{\frac{m v^{2}}{k}} = \sqrt{\frac{W v^{2}}{g k}} = v \sqrt{\frac{W}{k g}}$

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The maximum compression produced in a spring of force constant k, when a block of weight W moving with velocity v on a frictionless horizontal surface compresses it, shall be:

The correct option is C v√Wk.gFrom conservation of energy, KE lost by the block =PE gained by the spring so 12mv2=12kd2or d=√mv2k=√Wv2gk=v√Wkg

The maximum compression produced in a spring of force constant $k$ , when a block of weight $W$ moving with velocity $v$ on a frictionless horizontal surface compresses it, shall be:

The correct option is C $v \sqrt{\frac{W}{k . g}}$
From conservation of energy,
KE lost by the block $= P E$ gained by the spring
so $\frac{1}{2} m v^{2} = \frac{1}{2} k d^{2}$
or $d = \sqrt{\frac{m v^{2}}{k}} = \sqrt{\frac{W v^{2}}{g k}} = v \sqrt{\frac{W}{k g}}$