91Ó°ÊÓ

Elastic potential energy is the energy stored in an object when it is compressed or stretched. For a spring, this energy is given by the formula
U = \frac{1}{2} k x^2
Here:
- \( k \) is the spring constant, which measures the stiffness of the spring. In our exercise, it is 1960 N/m (converted from 19.6 N/cm).
- \( x \) is the distance the spring is compressed or stretched, which is 0.2 m (converted from 20.0 cm).
By substituting these values into the formula, we get: U = \frac{1}{2} \times 1960 \times (0.2)^2
So, the elastic potential energy stored in the compressed spring is 39.2 J.
This energy is crucial because when the spring is released, this stored energy can be converted into other forms of energy, like kinetic or gravitational potential energy.

Gravitational potential energy is the energy an object possesses due to its height above the ground. The formula to calculate this energy is:
\Delta U_g = mgh
Where:
- \( m \) is the mass of the object, here it is 2.00 kg.
- \( g \) is the acceleration due to gravity, which is approximately 9.8 m/s^2.
- \( h \) is the height change. For an incline, \( h \) can be expressed as \( h = d \sin (30.0°) \).
Since \( h \) is not directly given, we calculate it using the distance along the incline, \( d \), we need to find in the next step.

For our exercise, the change in gravitational potential energy is given by: \Delta U_g = 2.00 \times 9.8 \times d \sin (30.0°)
So, \Delta U_g = 9.8d J.
Here, our goal is to determine \( d \) which represents how far the block travels up the incline.

Energy conservation is a fundamental principle stating that energy cannot be created or destroyed, only converted from one form to another. In the context of our problem, the elastic potential energy stored in the compressed spring will convert entirely into gravitational potential energy as the block moves up the incline.
First, we found that the elastic potential energy of the spring is 39.2 J. According to the energy conservation principle, this energy will convert into gravitational potential energy when the block travels up the incline:
U = \Delta U_g
By equating the two, we have:
39.2 J = 9.8d J
Solving for \( d \), we have:
d = \frac{39.2}{9.8} = 4.00 m
This result tells us the block will travel a distance of 4.00 meters along the incline.
Understanding these concepts helps us apply energy conservation principles to solve problems involving the conversion of elastic potential energy to gravitational potential energy in real-world physics scenarios.