Question

0.2 kg mass gets pushed down against a vertical spring with spring constant k=1000N/m. when the spring is compressed by 10 cm the object is let go. what is the speed of the object the second its let go from the spring. (i.e. the spring is at its equilibrium length) answer in m/s

Answer 1

Using Energy conservation:

KEi + PEi = KEf + PEf

KEi = 0, since initially object is at rest

PEi = (1/2)*k*x^2 = Initial spring potential energy due to compresion of spring

x = initial compression = 10 cm = 0.10 m

KEf = (1/2)*m*V^2 = final KE

V = speed of object when spring is at equilibrium = ?

PEf = 0, since finally compression in spring is zero

So,

0 + (1/2)*k*x^2 = (1/2)*m*V^2

V = x*sqrt (k/m)

k = Spring constant = 1000 N/m

m = mass of object = 0.2 kg

Using given values

V = 0.1*sqrt (1000/0.2)

V = 7.07 m/sec

Let me know if you've any query.