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A particle is projected vertically upward with speed u = 10m/s. During its journey air applies a force of -0.2v^2 on the particle. what is the maximum height attained by the particle? (m =2kg). a) 5 ln(2) b) 3 ln(2). c) 2 ln(2) d) 10 ln(2)?
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Problem: Find the maximum height attained by a particle projected vertically upward with speed u = 10m/s, considering the force applied by the air (-0.2v^2) during its journey. (m = 2kg)
Solution:
We can solve this problem using the principle of conservation of energy. According to this principle, the total energy of a system remains constant if no external forces act on it.
The total energy of the particle at any point during its journey can be expressed as the sum of its kinetic energy (KE) and potential energy (PE).
Total energy (E) = KE + PE
Initially, the particle is at the ground level with zero potential energy and has a kinetic energy of:
KE = 1/2 * m * u^2 = 1/2 * 2 * 10^2 = 100 J
At the maximum height (h) attained by the particle, its velocity (v) becomes zero, and its kinetic energy is zero. Therefore, the total energy at this point is equal to its potential energy.
Total energy (E) = PE = mgh
where g is the acceleration due to gravity.
During the journey, the air applies a force of -0.2v^2 on the particle, which opposes its motion. This force is a non-conservative force and does not obey the principle of conservation of energy. Therefore, we need to consider it separately.
The work done by the air force (W) can be calculated as the negative of the change in the kinetic energy of the particle.
W = -ΔKE = -[KE(final) - KE(initial)]
At the maximum height, the velocity of the particle is zero, and its KE is zero. Therefore, the work done by the air force is equal to the initial kinetic energy of the particle.
W = -KE(initial) = -100 J
The negative sign indicates that the work done by the air force is against the direction of motion of the particle.
The work done by a non-conservative force is equal to the loss in the mechanical energy of the system. Therefore, the total energy of the particle at the maximum height is given by:
E = KE(initial) + W = 0 J
Substituting the values, we get:
mgh = 0 J - 100 J
2 * 9.81 * h = -100
h = -100 / (2 * 9.81) = -5.10 m
The negative value of h indicates that the particle does not reach the maximum height. It falls back to the ground due to the air resistance.
To find the time taken by the particle to reach the maximum height, we can use the equation of motion:
v = u + gt
At the maximum height, v = 0 and u = 10 m/s. Therefore, t = u/g = 10/9.81 = 1.02 s
During this time, the particle travels a distance of:
s = ut + 1/2 * gt^2 = 10 * 1.02 - 1/2 * 9.81 * 1.02^2 = 5.10 m
Therefore, the maximum height attained by the particle is zero. It falls back to the ground due to the air resistance.
Answer: (c) 2 ln(2)
Solution:
We can solve this problem using the principle of conservation of energy. According to this principle, the total energy of a system remains constant if no external forces act on it.
The total energy of the particle at any point during its journey can be expressed as the sum of its kinetic energy (KE) and potential energy (PE).
Total energy (E) = KE + PE
Initially, the particle is at the ground level with zero potential energy and has a kinetic energy of:
KE = 1/2 * m * u^2 = 1/2 * 2 * 10^2 = 100 J
At the maximum height (h) attained by the particle, its velocity (v) becomes zero, and its kinetic energy is zero. Therefore, the total energy at this point is equal to its potential energy.
Total energy (E) = PE = mgh
where g is the acceleration due to gravity.
During the journey, the air applies a force of -0.2v^2 on the particle, which opposes its motion. This force is a non-conservative force and does not obey the principle of conservation of energy. Therefore, we need to consider it separately.
The work done by the air force (W) can be calculated as the negative of the change in the kinetic energy of the particle.
W = -ΔKE = -[KE(final) - KE(initial)]
At the maximum height, the velocity of the particle is zero, and its KE is zero. Therefore, the work done by the air force is equal to the initial kinetic energy of the particle.
W = -KE(initial) = -100 J
The negative sign indicates that the work done by the air force is against the direction of motion of the particle.
The work done by a non-conservative force is equal to the loss in the mechanical energy of the system. Therefore, the total energy of the particle at the maximum height is given by:
E = KE(initial) + W = 0 J
Substituting the values, we get:
mgh = 0 J - 100 J
2 * 9.81 * h = -100
h = -100 / (2 * 9.81) = -5.10 m
The negative value of h indicates that the particle does not reach the maximum height. It falls back to the ground due to the air resistance.
To find the time taken by the particle to reach the maximum height, we can use the equation of motion:
v = u + gt
At the maximum height, v = 0 and u = 10 m/s. Therefore, t = u/g = 10/9.81 = 1.02 s
During this time, the particle travels a distance of:
s = ut + 1/2 * gt^2 = 10 * 1.02 - 1/2 * 9.81 * 1.02^2 = 5.10 m
Therefore, the maximum height attained by the particle is zero. It falls back to the ground due to the air resistance.
Answer: (c) 2 ln(2)
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A particle is projected vertically upward with speed u = 10m/s. During its journey air applies a force of -0.2v^2 on the particle. what is the maximum height attained by the particle? (m =2kg). a) 5 ln(2) b) 3 ln(2). c) 2 ln(2) d) 10 ln(2)?
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A particle is projected vertically upward with speed u = 10m/s. During its journey air applies a force of -0.2v^2 on the particle. what is the maximum height attained by the particle? (m =2kg). a) 5 ln(2) b) 3 ln(2). c) 2 ln(2) d) 10 ln(2)? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A particle is projected vertically upward with speed u = 10m/s. During its journey air applies a force of -0.2v^2 on the particle. what is the maximum height attained by the particle? (m =2kg). a) 5 ln(2) b) 3 ln(2). c) 2 ln(2) d) 10 ln(2)? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle is projected vertically upward with speed u = 10m/s. During its journey air applies a force of -0.2v^2 on the particle. what is the maximum height attained by the particle? (m =2kg). a) 5 ln(2) b) 3 ln(2). c) 2 ln(2) d) 10 ln(2)?.
A particle is projected vertically upward with speed u = 10m/s. During its journey air applies a force of -0.2v^2 on the particle. what is the maximum height attained by the particle? (m =2kg). a) 5 ln(2) b) 3 ln(2). c) 2 ln(2) d) 10 ln(2)? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A particle is projected vertically upward with speed u = 10m/s. During its journey air applies a force of -0.2v^2 on the particle. what is the maximum height attained by the particle? (m =2kg). a) 5 ln(2) b) 3 ln(2). c) 2 ln(2) d) 10 ln(2)? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle is projected vertically upward with speed u = 10m/s. During its journey air applies a force of -0.2v^2 on the particle. what is the maximum height attained by the particle? (m =2kg). a) 5 ln(2) b) 3 ln(2). c) 2 ln(2) d) 10 ln(2)?.
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