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A 1.10 kg block is attached to a spring with spring constant 17 N/m. While the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 46 cm/s.
A) What is the amplitude of the subsequent oscillations?
B) What is the block's speed at the point where x = 0.25 A?

Sagot :

Answer:

Explanation:

The kinetic energy of block will be converted into potential energy of spring .

If A be the amplitude of oscillations

1 /2 k A² = 1/2 m v²

17 A² = 1.1 x .46²

A² = .0137

A= 11.7 cm

B )

when x = .25 A = .25 x 11.7 = 2.9 cm

potential energy = 1/2 k x²

= .5 x 17 x ( .029 )² = .00715 J

kinetic energy = 1/2 m v²

1/2 m v²  +  .00715  = .5 1.1 x .46²

1/2 m v²  +  .00715  = .1164

1/2 m v² = .10925

.5 x 1.1 x v²=  .10925

v² = .1986

v = .4456 m /s

= 44.56 cm /s

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