# Hooke's Law With an Inclined plane?

Suppose you have a helical spring attached to the top of an inclined plane and there was a mass on the spring. And they asked you to find the work done or the energy held in the spring. please explain how to do this :/

### 1 Answer

- electron1Lv 79 years agoFavorite Answer
As the object slides down an inclined plane, its potential energy decreases. As the spring stretches its potential energy increases. Neglecting friction, the increase of the spring’s potential energy is equal to the decrease of the object’s potential energy.

Increase of PE of spring = ½ * k * d^2

Decrease of PE of object = mass * g * decrease of height

The object’s potential energy is dependent on the vertical distance that the object moves as it slides down the inclined plane.

The distance, which the object moves down the plane, is represented by the length of the inclined plane. The vertical distance is represented by the height of the inclined plane. The length is the hypotenuse of a right triangle. The height is the side opposite the angle.

Sin θ = side opposite/ hypotenuse = vertical distance/ distance moved down the plane

Sin θ = decrease of height/ distance moved down the plane

decrease of height = sin θ * distance moved down the plane

Decrease of PE of object = mass * g * sin θ * d

Increase of PE of spring = ½ * k * d^2

Increase of PE of spring = Decrease of PE of object

½ * k * d^2 = m * g * h

h = sin θ * d

½ * k * d^2 = m * g * sin θ * d

Divide both sides by d

½ * k * d = m * g * sin θ

Divide both sides by ½ * k

d = (2 * m * g * sin θ) ÷ k

Potential energy of spring = ½ * k * d^2

Potential energy of spring = ½ * k * [(2 * m * g * sin θ) ÷ k]^2

Example:

The spring constant = 490 N/m

The angle of the inclined plane = 30˚

The mass of the object = 5 kg

d = (2 * m * g * sin θ) ÷ k = (2 * 5 * 9.8 * sin 30) ÷ 490

d = 0.1 meter

Potential energy of spring = ½ * 490 * 0.1^2

Potential energy of spring = 2.45 J