Question

Leash a block of 0.5 kg down the entire length of an inclined plane forming a name of 37 ° with the horizontal. Friction with the plane is negligible. upon leaving the inclined plane, the block compresses a spring a spring on a polished horizontal surface. if the length of the plane is 2.5m and if the spring is and compressed by 60 cm what is the elasticity constant of the spring?

Answer 1

The potential energy of the block is given by:
V = m*g*h
m mass
g = 9.81m/s²
h height

The potential energy of a spring is given by:
V = 0.5 * k * x²

k spring constant
x compression of the spring

If the block starts from rest it has potential energy, but no kinetic energy. As it slides down the incline potential energy is converted into kinetic energy. When the block hits the spring the kinetic energy is converted into spring's potential energy. If the spring is fully compressed and the block is at rest again, the block has transferred all its energy into the spring. No energy is lost. So we can write:

m * g * h = 0.5 * k * x²

m = 0.5 kg
g = 9.81 m/s²
h = 2.5m * sin 37° = 1,5 m
x = 0,6 m

Solve for k.

k = 2 * m * g * h / x² = 40.8 N/m