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A ball of mass 0.2 kg is thrown vertically upwards by applying a force by hand. If the hand moves 0.2 m while applying the force and the ball goes upto 2 m height further, find the magnitude of the force. Consider g = 10 m/s2
[AIEEE 2006]
- a)4 N
- b)16 N
- c)20 N
- d)22 N
Correct answer is option 'D'. Can you explain this answer?
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A ball of mass 0.2 kg is thrown vertically upwards by applying a force...
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A ball of mass 0.2 kg is thrown vertically upwards by applying a force...
**Given:**
- Mass of the ball, m = 0.2 kg
- Distance covered by the hand while applying the force, s = 0.2 m
- Height reached by the ball, h = 2 m
- Acceleration due to gravity, g = 10 m/s²
**To find:**
The magnitude of the force applied to the ball.
**Assumption:**
- Neglecting air resistance and any other external forces.
**Explanation:**
1. We can use the work-energy principle to solve this problem. According to the principle, the work done on an object is equal to the change in its kinetic energy.
2. The work done on the ball can be calculated using the formula: Work = Force × Distance
3. The work done on the ball is equal to the change in its potential energy. Since the ball is thrown vertically upwards, the change in potential energy is given by: ΔPE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height reached by the ball.
4. The work done on the ball is also equal to the force applied multiplied by the distance covered by the hand: Work = Force × Distance
5. Equating the two expressions for work, we have: Force × Distance = mgh
6. Substituting the given values: Force × 0.2 = 0.2 × 10 × 2
7. Solving for the force, we get: Force = (0.2 × 10 × 2) / 0.2 = 20 N
**Therefore, the magnitude of the force applied to the ball is 20 N, which corresponds to option 'D'.
- Mass of the ball, m = 0.2 kg
- Distance covered by the hand while applying the force, s = 0.2 m
- Height reached by the ball, h = 2 m
- Acceleration due to gravity, g = 10 m/s²
**To find:**
The magnitude of the force applied to the ball.
**Assumption:**
- Neglecting air resistance and any other external forces.
**Explanation:**
1. We can use the work-energy principle to solve this problem. According to the principle, the work done on an object is equal to the change in its kinetic energy.
2. The work done on the ball can be calculated using the formula: Work = Force × Distance
3. The work done on the ball is equal to the change in its potential energy. Since the ball is thrown vertically upwards, the change in potential energy is given by: ΔPE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height reached by the ball.
4. The work done on the ball is also equal to the force applied multiplied by the distance covered by the hand: Work = Force × Distance
5. Equating the two expressions for work, we have: Force × Distance = mgh
6. Substituting the given values: Force × 0.2 = 0.2 × 10 × 2
7. Solving for the force, we get: Force = (0.2 × 10 × 2) / 0.2 = 20 N
**Therefore, the magnitude of the force applied to the ball is 20 N, which corresponds to option 'D'.
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A ball of mass 0.2 kg is thrown vertically upwards by applying a force...
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A ball of mass 0.2 kg is thrown vertically upwards by applying a force by hand. If the hand moves 0.2 m while applying the force and the ball goes upto 2 m height further, find the magnitude of the force. Consider g = 10 m/s2[AIEEE 2006]a)4 Nb)16 Nc)20 Nd)22 NCorrect answer is option 'D'. Can you explain this answer?
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