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Step-by-Step Solution
Step 1: Identify the Known Quantities
• Mass of the ball, $m = 0.2 \, \text{kg}$
• Acceleration due to gravity, $g = 10 \, \text{m/s}^2$
• Distance over which the force is applied by hand, $d_{hand} = 0.2 \, \text{m}$
• Additional height gained after leaving the hand, $h = 2 \, \text{m}$
• Total upward distance during which gravity acts, $d_{total} = d_{hand} + h = 0.2 + 2 = 2.2 \, \text{m}$
Step 2: State the Work-Energy Principle
According to the work-energy theorem, the total work done on the ball equals its change in kinetic energy ($\Delta K$). Initially, the ball is at rest, and finally (at the top of its flight) it comes to rest again, so its change in kinetic energy is $0$:
$W_{total} = \Delta K = 0$.
Step 3: Express the Total Work Done
Two forces do work on the ball:
Force by the hand, $F$, acting over $0.2 \, \text{m}$
Gravitational force, $mg$, acting downward over $2.2 \, \text{m}$
Thus,
$W_{\text{hand}} + W_{\text{gravity}} = 0$.
Step 4: Calculate the Work Done by Each Force
Work done by the hand: $W_{\text{hand}} = F \times d_{hand} = F \times 0.2$.
Work done by gravity: Since gravity acts downwards against the ball’s upward displacement, the work done by gravity is $W_{\text{gravity}} = -mg \times d_{total} = - (0.2 \times 10) \times 2.2.$
Note the negative sign indicates gravity opposes the motion.
Step 5: Apply the Work-Energy Theorem
Since the net work done is zero,
$F \times 0.2 + \Bigl[-(0.2 \times 10 \times 2.2)\Bigr] = 0.$
Simplifying,
$0.2F \;-\; (0.2 \times 10 \times 2.2) = 0.$
$0.2F - (0.2 \cdot 10 \cdot 2.2) = 0.$
$0.2F - 4.4 = 0.$
$0.2F = 4.4.$
$F = \frac{4.4}{0.2} = 22 \, \text{N}.$
Step 6: Conclude the Answer
Hence, the magnitude of the force applied by the hand is $22 \, \text{N}.$
