Question

A 1.10kg block is attached to a spring with spring constant 18 N/m. While the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 40cm/s .What is the block's speed at the point where x = 0.45 A?

Answer 1

The velocity is
`v_x = 0.356 \ m/s`

Step-by-step explanation:

From the question we are told that

The mass is
`m = 1.10 \ kg`

The spring constant is
`k = 18 \ N/m`

The speed is
`v = 40 \ cm / s = 0.4 m/s`

The position considered is x = 0.45 A

Here A is the amplitude which is mathematically represented as

`A = v * \sqrt{(m)/(k) }`

=>
`A = 0.4 * \sqrt{(1.10)/(18 ) }`

=>
`A = 0.0989 \ m`

So
`x = 0.45 * 0.0989`

=>
`x = 0.045 \ m`

Generally the speed at x is mathematically represented as

`v_x = \sqrt{ (k)/(m) * [A^2 - x^2 ]}`

=>
`v_x = \sqrt{ (18)/( 1.10) * [0.0989^2 - 0.045^2 ]}`

=>
`v_x = 0.356 \ m/s`