If a 0.4 kg ball is dropped from a height of 7 m, what is its kinetic energy when it hits the ground?
Mgh=(0.4kg)*(9.8m/s^2)*(7m)=?
To find the kinetic energy of the ball when it hits the ground, you need to know its velocity just before impact.
To find the velocity, we can use the principle of conservation of energy. The potential energy of the ball at the starting point (7 m height) will convert into kinetic energy as it falls.
The potential energy is given by the formula:
Potential Energy = m * g * h
Where
m = mass of the ball (0.4 kg)
g = acceleration due to gravity (approximately 9.8 m/s²)
h = height (7 m)
Substituting the given values into the formula, we get:
Potential Energy = 0.4 kg * 9.8 m/s² * 7 m
Now, we know that the initial potential energy will be converted into kinetic energy before it reaches the ground.
The formula for kinetic energy is:
Kinetic Energy = 0.5 * m * v²
Where
m = mass of the ball (0.4 kg)
v = velocity of the ball
Since the ball starts from rest at the top, its initial velocity is 0 m/s. Therefore, the kinetic energy formula simplifies to:
Kinetic Energy = 0.5 * m * v²
= 0.5 * 0.4 kg * (v²)
We can now solve for the velocity:
Potential Energy = Kinetic Energy
0.4 kg * 9.8 m/s² * 7 m = 0.5 * 0.4 kg * (v²)
Simplifying the equation:
27.44 J = 0.2 kg * (v²)
27.44 J = 0.2 kg * v²
Now, solve for v²:
v² = 27.44 J / 0.2 kg
v² = 137.2 m²/s²
Finally, take the square root to find the velocity:
v = √(137.2 m²/s²)
v ≈ 11.70 m/s
Now that we know the velocity, we can substitute it back into the kinetic energy formula to find the answer:
Kinetic Energy = 0.5 * m * v²
Kinetic Energy = 0.5 * 0.4 kg * (11.70 m/s)²
Calculating the value:
Kinetic Energy ≈ 32.53 J
Therefore, the kinetic energy of the ball when it hits the ground is approximately 32.53 Joules.