Question

When a body of mass 0.25 kg is attached to a vertical massless spring, it is extended 5.0 cm from its unstretched length of 4.0 cm. The body and spring are placed on a horizontal frictionless surface and rotated about the held end of the spring at 2.0 rev/s. How far is the spring stretched?

Answer 1

`d=0.165m`

Step-by-step explanation:

Given

`m=0.25kg`,
`x_(1)=5cm*(1m)/(100cm)=0.05m`,
`x_(2)=4cm*(1m)/(100cm)=0.04m`,
`v=2(rev)/(s)`

The tension of the spring is

`F_(k)=K*x_(1)=m*g`

`K=(m*g)/(x_(1))`

`K=(0.25kg*9.8m/s^2)/(0.05m)=49N/m`

The force in the spring is equal to centripetal force so

`F_(c)=(m*v^2)/(r)`

`v=w*r=2\pi*r`

But Fc is also

Fc=KxΔr

`F_(c)=K*(r-x_(2))`

Replacing

`m*4\pi^2*r=K*(r-x_(2))`

`0.25kg*4\pi^2*r=49*(r-0.04m)`

`r=0.205m`

total distance is

`d=0.205-0.04=0.165m`