Amass of 2 kilograms is on a spring with spring constant k newtons per meter with no damping. suppose the system is at rest and at time t = 0 the mass is kicked and starts traveling at 2 meters per second. how large does k have to be to so that the mass does not go further than 3 meters from the rest position?

k = 0.445

Explanation: F=kx (x being the extension which is 3m in this case)

We need value of F to find k.

F = ma (m=2kg)

To find F we need a

Since the mass starts travelling at u=2m/s, for the mass to stop at 3m, the final velocity (v) should be equal to zero.

Using equation of motion: v^2 - u^2 = 2as

0 - 4 = 2*a*3

Solving for a leads to: a = -2/3 m/s^2

Hence the deceleration must be 0.667 m/s^2

Using a to find Force: F = 2*0.667 = 1.334

Using F to find k: k = F/x = 1.334/3 = 0.445