A 4.00-kg mass is connected to a cable attached to a spring with a spring constant of 75.0 N/m. when the spring is at equlibrium, the mass has a speed of 2.00 m/s. what is the speed of the mass when it has fallen 1.25 m?

A spring is oriented vertically and compressed by a distance of 0.098. It has a spring constant of 5 100 N/m. A block with a mass 0.255 kg is placed at rest on the upper end of the sprng. When the block is released, it moves upward and leaves the spring. To what maximum height above the point of release does it rise? (Round to the nearest 2 decimals)

The problem uses the ideal spring force, in one dimension.  F = -kx, where x is the deviation from equilibrium, and k is the spring constant.  Calculate the particle's maximum speed (vmax), with these numbers: k = 2000 N/m = 2000 kg/s2 m = 5 kg (the mass of the particle) x0 = 0.2 m (the initial position) v0 = 5 m/s (the initial velocity) The all-too-common erroneous solution: v max = A ω = x 0 k / m = 0.2 m ⋅ 20 / s = 4 m/s . Your problem: You may not understand this problem fully, and may not be able to solve it.  Nevertheless, spot the egregious wrongness of the answer.

A spring has a force constant of 18.9 N/m and a rest length of 0.35 m.The spring is stretched to a length of 1.2 m and attached to a mass of 0.45 kg. When the spring is released, what is the speed of the mass when the spring has contracted to 0.70 m long?