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Karnataka Board PUCPUC Science Class 11

A block of mass 5 kg is suspended from the end of a vertical spring which is stretched by 10 cm under the load of the block. The block is given a sharp impulse from below, so that it acquires an - Physics

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Question

A block of mass 5 kg is suspended from the end of a vertical spring which is stretched by 10 cm under the load of the block. The block is given a sharp impulse from below, so that it acquires an upward speed of 2 m/s. How high will it rise? Take g = 10 m/s2

Numerical

Solution

Given,

Mass of the block, m = 5 kg

Compression in the string with the load, x = 10 cm = 0 . 1 m

Initial speed in upward direction, ν = 2 m/s,

h = ?, g = 10 m sec2

So, F = kx = mg

⇒`k = (mg)/x`

⇒ `50/0.1 = 500` N/m

Total energy just after the impulse,

`E = 1/2 mv^2 + 1/2 kx^2`   ...(i)

Total energy at a height h

= `1/2 k(h - x)^2 + mgh`

On solving, we get:

h = 0.2 m

h = 20 cm

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Chapter 8: Work and Energy - Exercise [Page 135]

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HC Verma Concepts of Physics Vol. 1 [English] Class 11 and 12
Chapter 8 Work and Energy
Exercise | Q 41 | Page 135

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