The initial speed of the rock is 30.32 m/s
Step-by-step explanation:
The work-energy theorem is
1. work = change in the kinetic energy
2. work = force × distance
2. K.E = [tex]\frac{1}{2}[/tex] m v²
3. [tex]F*d=\frac{1}{2}m[(v_{f})^{2}-(v_{i})^{2}][/tex], where m is the mass,
[tex]v_{f}[/tex] is the final velocity and [tex]v_{i}[/tex] is the initial velocity
A 3.00 N rock is throw vertically into the air from the ground
At h = 15.0 m, v = 25 m/s upward
We need to find the initial speed of the rock
∵ The weight of the rock = 3 N
∵ Weight = m g, where m is the mass and g is the acceleration
gravity (-9.8 m/s²)
∴ 3 = m (-9.8)
∵ [tex]F*d=\frac{1}{2}m[(v_{f})^{2}-(v_{i})^{2}][/tex]
∵ F = (-9.8)m , d = 15 m , [tex]v_{f}[/tex] = 25 m/s
- Substitute these values in the rule above
∴ [tex](-9.8)m(15)=\frac{1}{2}m[(25)^{2}-(v_{i})^{2}][/tex]
- Divide both sides by m
∴ [tex](-9.8)(15)=\frac{1}{2}[(25)^{2}-(v_{i})^{2}][/tex]
- Multiply both sides by 2
∴ -294 = 625 - ([tex]v_{i}[/tex])²
- Add ([tex]v_{i}[/tex])² to both sides
∴ ([tex]v_{i}[/tex])² - 294 = 625
- Add 294 from both sides
∴ ([tex]v_{i}[/tex])² = 919
- Take √ for both sides
∴ [tex]v_{i}[/tex] = 30.32 m/s
The initial speed of the rock is 30.32 m/s
Learn more:
You can learn more about word problem in brainly.com/question/4586309
#LearnwithBrainly
