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Standard VII

Physics

Law of Conservation of Energy

A block of ma...

Question

A block of mass $250 g$ is kept (does not stick to spring) on a vertical spring of spring constant $100 N / m$ fixed from below. The spring is now compressed to have a length $10 c m$ shorter than its natural length and the system is released from this position. How high does the block rise from this position? Take $g = 10 m / s^{2}$ .

20 c m

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30 c m

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40 c m

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50 c m

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Solution

The correct option is A $20 c m$
PE held by KE= $k x^{2} / 2 = 100 (0.1)^{2} / 2 = 0.5 J$
Hence, KE transferred to the mass= $0.5 J = 0.5 m u^{2}$
$u = \sqrt{1 / 0.25} = 2 m / s$
$v^{2} = u^{2} + 2 a s$
$0 = 4 + 2 (- 10) s$
$s = 0.2 m = 20 c m$

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Similar questions

Q. A block of mass

250 g

is kept on a vertical spring of spring constant

100 N / m

fixed from below. The spring is now compressed to have a length

10 c m

shorter than its natural length and the system is released from this position. How high does the block rise? (

g = 10 m / s^{2}

)

Q. A block of mass

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fixed at the other end and kept on smooth level ground. The block is pulled by a distance of

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\frac{1}{2} cm

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.A block of mass

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A collar of mass 10 kg is released from rest from position A, when spring is in natural length and slides along a rough vertical fixed rod. The block comes to rest at point B, 4 m below the point A. The spring constant of spring is 100 N/m and natural length of spring is 3 m. Find work done by friction (‘O’ is a fixed point)

Q. The pulley shown in figure (10-E11) has a radius of 20 cm and moment of inertia 0⋅2 kg-m². The string going over it is attached at one end to a vertical spring of spring constant 50 N/m fixed from below, and supports a 1 kg mass at the other end. The system is released from rest with the spring at its natural length. Find the speed of the block when it has descended through 10 cm. Take g = 10 m/s².

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The correct option is A 20 cmPE held by KE=kx2/2=100(0.1)2/2=0.5JHence, KE transferred to the mass= 0.5J=0.5mu2u=√1/0.25=2m/sv2=u2+2as0=4+2(−10)ss=0.2m=20cm

The correct option is A $20 c m$
PE held by KE= $k x^{2} / 2 = 100 (0.1)^{2} / 2 = 0.5 J$
Hence, KE transferred to the mass= $0.5 J = 0.5 m u^{2}$
$u = \sqrt{1 / 0.25} = 2 m / s$
$v^{2} = u^{2} + 2 a s$
$0 = 4 + 2 (- 10) s$
$s = 0.2 m = 20 c m$