To solve the problem, we need to determine how high the rock will rise when it's launched by the spring. Here's a detailed, step-by-step solution:
1. Convert the compression distance to meters:
The given compression distance is 32 centimeters. Converting this to meters:
[tex]\[
32 \text{ cm} = 0.32 \text{ m}
\][/tex]
2. Calculate the potential energy stored in the spring (Elastic Potential Energy):
The potential energy [tex]\( E \)[/tex] stored in a compressed or stretched spring can be calculated using Hooke’s law:
[tex]\[
E = \frac{1}{2} k x^2
\][/tex]
where:
- [tex]\( k \)[/tex] is the spring constant (2500 [tex]\( \frac{N}{m} \)[/tex])
- [tex]\( x \)[/tex] is the compression distance (0.32 m)
Substituting the given values:
[tex]\[
E = \frac{1}{2} \times 2500 \times (0.32)^2
\][/tex]
Performing the multiplication:
[tex]\[
E = \frac{1}{2} \times 2500 \times 0.1024
\][/tex]
[tex]\[
E = 1250 \times 0.1024
\][/tex]
[tex]\[
E = 128 \, \text{Joules}
\][/tex]
3. Calculate the height the rock will rise (Gravitational Potential Energy):
The gravitational potential energy [tex]\( E \)[/tex] at a height [tex]\( h \)[/tex] is given by:
[tex]\[
E = mgh
\][/tex]
where:
- [tex]\( m \)[/tex] is the mass of the rock (1.5 kg)
- [tex]\( g \)[/tex] is the acceleration due to gravity (9.81 [tex]\( \frac{m}{s^2} \)[/tex])
- [tex]\( h \)[/tex] is the height the rock will rise
We set the potential energy stored in the spring equal to the gravitational potential energy:
[tex]\[
\frac{1}{2} k x^2 = mgh
\][/tex]
[tex]\[
128 = 1.5 \times 9.81 \times h
\][/tex]
Solving for [tex]\( h \)[/tex]:
[tex]\[
h = \frac{128}{1.5 \times 9.81}
\][/tex]
[tex]\[
h = \frac{128}{14.715}
\][/tex]
[tex]\[
h \approx 8.70 \text{ m}
\][/tex]
Therefore, the height that the rock will rise is approximately 8.70 meters. Thus, the closest given answer choice is 9 m.
