A bowling ball that weighs 70N is dropped from a tower 15m above the ground. Disregarding air friction, what is the kinetic energy (KE) of the ball when it reaches the ground?
1050 J
Ah, the ball's going on a gravity-induced journey! Let's calculate its kinetic energy when it reaches the ground, shall we?
Now, the formula for kinetic energy (KE) is KE = 1/2 * m * v^2. Since the ball is dropped, we know its initial velocity (v) is 0.
To find the mass (m) of the ball, we divide its weight (70N) by the acceleration due to gravity (9.8 m/s^2). So, m = 70N / 9.8 m/s^2.
Plugging all the numbers into the equation, we have KE = 1/2 * (70N / 9.8 m/s^2) * (0 m/s)^2.
Now, since anything multiplied by zero is zero, we can conclude that the ball's kinetic energy when it reaches the ground is zero. It's as still as a scared cat trying to avoid water!
To calculate the kinetic energy (KE) of the bowling ball when it reaches the ground, we can use the formula:
KE = 1/2 * m * v^2
where
- KE is the kinetic energy
- m is the mass of the object
- v is the velocity of the object
Given:
- The weight of the bowling ball is 70N.
To find the mass (m) of the bowling ball, we can divide its weight by the acceleration due to gravity (g).
Weight (W) = m * g
Therefore, m = W / g
Given:
- The acceleration due to gravity is approximately 9.8 m/s^2.
So, m = 70 N / 9.8 m/s^2
Now, we can calculate the mass of the bowling ball.
m = 7.14 kg (approximately)
Next, we need to find the velocity (v) of the bowling ball when it reaches the ground. We can use the equation:
v^2 = u^2 + 2gh
where
- v is the final velocity
- u is the initial velocity (usually 0 for objects dropped)
- g is the acceleration due to gravity
- h is the height
Given:
- Initial velocity (u) is 0 m/s (since the ball is dropped)
- Acceleration due to gravity (g) is 9.8 m/s^2
- Height (h) is 15 m
Using the equation, we can solve for v.
v^2 = 0 + 2 * 9.8 m/s^2 * 15 m
v^2 = 294 m^2/s^2
Taking the square root of both sides, we find:
v = √294 m/s
Now, we can substitute the values of mass (m) and velocity (v) in the kinetic energy formula.
KE = 1/2 * 7.14 kg * (√294 m/s)^2
KE ≈ 1/2 * 7.14 kg * 294 m^2/s^2
KE ≈ 1036.38 J (approximately)
Therefore, the kinetic energy of the bowling ball when it reaches the ground is approximately 1036.38 Joules.
To calculate the kinetic energy (KE) of the bowling ball when it reaches the ground, we need to use the formula: KE = 1/2 mv^2, where m is the mass of the object and v is its velocity.
Since we are given the weight of the bowling ball (70N), we can use Newton's second law of motion, which states that weight (W) is equal to mass (m) multiplied by the acceleration due to gravity (g). Therefore, we can find the mass of the bowling ball using the formula: W = mg.
Given that the weight of the bowling ball is 70N and the acceleration due to gravity is approximately 9.8 m/s^2, we can calculate the mass (m) as follows:
70N = m * 9.8 m/s^2
Solving for m, we get:
m = 70N / 9.8 m/s^2
m ≈ 7.14 kg
Now that we know the mass of the ball, we can calculate its velocity (v) when it reaches the ground using the equation: v = √(2gh), where g is the acceleration due to gravity (9.8 m/s^2) and h is the height (15m) from which the ball was dropped.
Substituting the given values, we have:
v = √(2 * 9.8 m/s^2 * 15m)
v ≈ √(294 m^2/s^2)
v ≈ 17.15 m/s
Finally, substituting the calculated values for mass (m ≈ 7.14 kg) and velocity (v ≈ 17.15 m/s) into the formula for kinetic energy (KE = 1/2 mv^2), we can find the kinetic energy as follows:
KE = 1/2 * 7.14 kg * (17.15 m/s)^2
Calculating this expression, we get:
KE ≈ 1/2 * 7.14 kg * 294.22 m^2/s^2
KE ≈ 1047.92 J
Therefore, the kinetic energy of the bowling ball when it reaches the ground is approximately 1047.92 Joules (J).
Final Kinetic Energy = Initial Potential energy
= Weight x Height
= 105 J