Question

A spring is attached to an inclined plane as shown in the figure. A block of mass m = 2.75 kg is placed on the incline at a distance d = 0.294 m along the incline from the end of the spring. The block is given a quick shove and moves down the incline with an initial speed v = 0.750 m/s. The incline angle is θ = 20.0°, the spring constant is k = 450 N/m, and we can assume the surface is frictionless. By what distance (in m) is the spring compressed when the block momentarily comes to rest?

Answer 1

Gravitational acceleration = g = 9.81 m/s2

Mass of the block = m = 2.75 kg

Angle of incline = = 20 degrees

Force constant of the spring = k = 450 N/m

Initial speed of the block = V = 0.75 m/s

Distance of the block from the spring = d = 0.294 m

Amount the spring is compressed when the block comes to rest = X

Height through which the block falls = H

H = (d + X)Sin

By conservation of energy the initial potential and kinetic energy of the block is converted into the potential energy of the spring as the block momentarily comes to rest.

X = 0.147 m or -0.106 m

Distance cannot be negative.

X = 0.147 m

Compression of the spring when the block momentarily comes to rest = 0.147 m