Final answer:
The stone strikes the ground with a speed of 9.39 m/s.
Step-by-step explanation:
We can use the Principle of Conservation of Energy to determine the speed with which the stone strikes the ground. According to this principle, the total mechanical energy of the stone remains constant throughout its fall. Initially, the stone only has potential energy due to its position above the ground. At the moment it strikes the ground, all of this potential energy is transformed into kinetic energy. Since the mass of the stone is given as 25 g, we can first convert it to kilograms by dividing by 1000: 25 g = 0.025 kg.
The potential energy of the stone at a height of 6 m can be calculated using the formula P.E. = mgh, where m is the mass, g is the acceleration due to gravity (9.8 m/s^2), and h is the height. Plugging in the values, we get:
- P.E. = (0.025 kg)(9.8 m/s^2)(6 m) = 1.47 J
Since the total mechanical energy is conserved, the potential energy at the beginning is equal to the kinetic energy at the end. We can use the formula K.E. = (1/2)mv^2 to find the speed (v) with which the stone strikes the ground. Rearranging the formula and substituting the known values, we have:
- 1.47 J = (1/2)(0.025 kg)v^2
Simplifying the equation and solving for v:
- v^2 = (2)(1.47 J) / (0.025 kg) = 88.2 m^2/s^2
- v = sqrt(88.2 m^2/s^2) = 9.39 m/s
Therefore, the stone strikes the ground with a speed of 9.39 m/s.