A (10.25) kg bowling ball is hung on a (11.50) m long rope. It is then pulled back until the rope makes an angle of (39.0) degrees with the vertical and released. Find the speed of the ball is at the lowest point. Give your answer in m/s and with 3 significant figures.
To find the speed of the bowling ball at the lowest point, we can use the principle of conservation of mechanical energy. At the highest point, the ball has only potential energy, and at the lowest point, it has both potential and kinetic energy.
First, let's find the potential energy of the ball at the highest point. The potential energy is given by the formula:
Potential Energy = mass × acceleration due to gravity × height
The mass of the bowling ball is given as 10.25 kg, and the height is the length of the rope, which is 11.50 m. The acceleration due to gravity is approximately 9.8 m/s². Substituting these values into the formula, we get:
Potential Energy = 10.25 kg × 9.8 m/s² × 11.50 m = 1191.17 Joules
Since energy is conserved, the potential energy at the highest point is equal to the sum of the potential and kinetic energy at the lowest point. At the lowest point, the potential energy is zero (as it is a reference point), so the total energy is equal to the kinetic energy:
Total Energy = Kinetic Energy = 0.5 × mass × velocity²
We are trying to find the velocity, so let's rearrange the equation:
velocity² = (2 × Total Energy) / mass
Now, substitute the known values into the equation. The total energy is equal to the potential energy at the highest point (1191.17 Joules). The mass of the ball is 10.25 kg. Plugging these values in, we get:
velocity² = (2 × 1191.17 Joules) / 10.25 kg = 231.31 m²/s²
To find the velocity, take the square root of both sides:
velocity = √(231.31 m²/s²) = 15.20 m/s
Therefore, the speed of the ball at the lowest point is 15.20 m/s (rounded to 3 significant figures).
the ball is pulled back from the lowest point, and gains gravitational potential energy
this energy becomes kinetic energy when the ball is released
... it is maximum at the low point
m g h = 1/2 m v^2 ... v = √(2 g h)
use trig to find the height above the low point