Question

A spring with constant k = 78 N/m is at the base of a frictionless, 30.0°-inclined plane. A 0.50-kg block is pressed against the spring, compressing it 0.20 m from its equilibrium position. The block is then released. If the block is not attached to the spring, how far up the incline will it travel before it stops?

Answer 1

Step-by-step explanation:

The given data is as follows.

Spring constant (k) = 78 N/m,
`\theta = 30^(o)`

Mass of block (m) = 0.50 kg

According to the formula of energy conservation,

mgh sin
`\theta = (1)/(2)kx^(2)`

h =
`(1)/(2) * (kx^(2))/(mg Sin \theta)`

=
`(78 * 0.04)/(2 * 0.5 * 9.8 * 0.5)`

= 0.64 m

Thus, we can conclude that the distance traveled by the block is 0.64 m.