Amass of 2 kg is attached to a spring and placed on a horizontal surface. the spring has a spring constant of 20 n/m, and the spring is compressed 0.1 m past its natural length. if the mass is released from this compressed position, what is the speed of the mass as it passes the natural length of the spring?

The speed of the mass : v = 0.316 m/s

The energy used to press a spring is included as the potential energy

Can be formulated:

Ep= potential energy

k = spring constant

x = change in spring length

If the spring is released from its pressure, this potential energy will turn into kinetic energy

so applies the law of conservation of mechanical energy (Em)

Ek = Ep

A mass of 2 kg is attached to a spring, a spring constant of 20 N/m, and the spring is compressed 0.1 m past its natural length.

m = 2 kg

k = 20 N/m

x = 0.1 m

Hooke's law

Keywords : spring,mass, spring constant,compressed position

find the sum of the two horizontal vectors and the sum of the two vertical vectors

isoceles right angled triangle

distance = 150m

displ = root (50 ^2 + 100^2)

root(2500+10000)

root 125000