When a body of mass 1.0 kg is suspended from a certain light spring hanging vertically, its length increases by 5 cm

When a body of mass 1.0 kg is suspended from a certain light spring hanging vertically, its length increases by 5 cm. By suspending 2.0 kg block to the spring and if the block is pulled through 10 cm and released, the maximum velocity of it in m/s is approximately

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Expert Solution

Given that

1×g=Kx

3g=K(x+5)

1/3=x/x+5

x=2.5

10=K×2.5×100

K= 400N hm

Now we have to find ω , We have formula which is

ω =√K/m

ω =√400/3

Vmax = Aω

Putting the values and evaluating we will get

Vmax = 10 × 15²√400/3

Vmax = 1.15 m/s

Therefore the maximum velocity will be 1.15m /s

Maximum velocity (Vmax) is 1.15 m/s

Nikhil Thakur answered 1 year ago

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