A force of 10 N is used to lift a 600-g ball from the ground to a height of 1.8 m, when it is let go. What is the speed of the ball when it is let go??
when it let go? IF it was held still at the 1.8 m height, then the initialspeed is zero.
To find the speed of the ball when it is let go, we can use the principle of conservation of energy. The initial potential energy gained by lifting the ball is converted into kinetic energy at the top.
Step 1: Convert the mass of the ball from grams to kilograms:
- The mass of the ball is given as 600 g.
- 1 kilogram (kg) is equal to 1000 grams (g).
- Therefore, the mass of the ball in kilograms is 600 g ÷ 1000 g/kg = 0.6 kg.
Step 2: Calculate the potential energy gained when lifting the ball:
- The potential energy (PE) gained by lifting an object is given by the equation: PE = m*g*h,
where m is the mass, g is the acceleration due to gravity, and h is the height.
- In this case, the mass (m) is 0.6 kg, the acceleration due to gravity (g) is 9.8 m/s^2 (approximately), and the height (h) is 1.8 m.
- Therefore, the potential energy gained is PE = 0.6 kg * 9.8 m/s^2 * 1.8 m = 10.44 J (rounded to two decimal places).
Step 3: Calculate the final kinetic energy of the ball:
- The kinetic energy (KE) of an object is given by the equation: KE = (1/2) * m * v^2,
where m is the mass and v is the velocity (speed).
- Since the potential energy is converted entirely into kinetic energy at the top, the potential energy is equal to the kinetic energy.
- Therefore, 10.44 J = (1/2) * 0.6 kg * v^2.
- We can solve for v^2 by rearranging the equation: v^2 = (2 * 10.44 J) / 0.6 kg = 34.8 m^2/s^2.
- Finally, find v by taking the square root of both sides: v = √(34.8 m^2/s^2) ≈ 5.89 m/s (rounded to two decimal places).
The speed of the ball when it is let go is approximately 5.89 m/s.