To calculate the total initial energy of the system, we need to consider both the kinetic energy (KE) and the potential energy (PE) of the ball when it is tossed.
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Kinetic Energy (KE) is given by the formula: \[ KE = \frac{1}{2} m v^2 \] where \( m \) is the mass of the ball and \( v \) is the initial velocity.
Given:
- \( m = 1.2 , \text{kg} \)
- \( v = 5.0 , \text{m/s} \)
Substituting these values: \[ KE = \frac{1}{2} \times 1.2 , \text{kg} \times (5.0 , \text{m/s})^2 \] \[ KE = \frac{1}{2} \times 1.2 \times 25 \] \[ KE = 0.6 \times 25 = 15 , \text{J} \]
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Potential Energy (PE) is given by the formula: \[ PE = mgh \] where \( g \) is the acceleration due to gravity, and \( h \) is the height above the ground.
Given:
- \( g = 9.8 , \text{N/kg} \)
- \( h = 20.0 , \text{m} \)
Substituting these values: \[ PE = 1.2 , \text{kg} \times 9.8 , \text{N/kg} \times 20.0 , \text{m} \] \[ PE = 1.2 \times 9.8 \times 20 \] \[ PE = 1.2 \times 196 = 235.2 , \text{J} \]
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Total Initial Energy (TE) is the sum of KE and PE: \[ TE = KE + PE \] \[ TE = 15 , \text{J} + 235.2 , \text{J} = 250.2 , \text{J} \]
Since the question asks for the total initial energy of the system rounded to significant figures, we can summarize:
The total initial energy of the system is approximately 250 J.
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