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A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7N on a table with coefficient of kinetic friction = 0.1. Work done by friction in 10s and change in K.E. of the body in 10s are respectively
- a)246.9 J, 635 J
- b)200 J, 600 J
- c)295 J, 625 J
- d)500 J, 600 J
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
A body of mass 2 kg initially at rest moves under the action of an ap...
Given : m = 2kg, u = 0, F = 7N, μk = 0.1, t = 10s,
W = ?
Acceleration produced by applied force
Force of friction, f = μkR = μkmg
= 0.1 × 2 × 9.8 = 1.96 N
Retardation produced by friction
Net acceleration = a = a1 + a2 = 3.50 – 0.98
= 2.52 m/s2
Distance moved by the body in 10 seconds
Work done by the frictional force = – f ×s
= – 1.96 × 126 = – 246.9 J
Velocity at the end of 10 s is
v = u + at = 0 + 2.52 × 10 = 25.2 m/s
Final K.E. = 
Initial K.E. = 0
∴ Change in KE = 635 – 0 = 635 J
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A body of mass 2 kg initially at rest moves under the action of an ap...
Given data:
- Mass of the body (m) = 2 kg
- Applied force (F) = 7 N
- Coefficient of kinetic friction (μ) = 0.1
- Time duration (t) = 10 s
Calculating work done by friction:
The work done by friction can be calculated using the formula:
Work done by friction (Wf) = Force of friction (f) × Distance (d)
The force of friction can be determined using the formula:
Force of friction (f) = μ × Normal force (N)
The normal force can be calculated as the product of the mass and acceleration due to gravity:
Normal force (N) = m × g
Where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Now, substituting the values into the equations:
Normal force (N) = 2 kg × 9.8 m/s^2 = 19.6 N
Force of friction (f) = 0.1 × 19.6 N = 1.96 N
To calculate the distance, we can use the equation of motion:
Distance (d) = Initial velocity (u) × time (t) + (1/2) × acceleration (a) × time squared (t^2)
Since the body is initially at rest, the initial velocity (u) is 0 m/s.
The acceleration (a) can be calculated using Newton's second law:
Force (F) = mass (m) × acceleration (a)
Substituting the values:
7 N = 2 kg × a
a = 3.5 m/s^2
Now, substituting the values into the equation of motion:
Distance (d) = 0 m/s × 10 s + (1/2) × 3.5 m/s^2 × (10 s)^2
Distance (d) = 0 + (1/2) × 3.5 m/s^2 × 100 s^2
Distance (d) = 0 + 175 m
Distance (d) = 175 m
Now, substituting the values into the equation for work done by friction:
Work done by friction (Wf) = 1.96 N × 175 m
Work done by friction (Wf) = 343 J
Therefore, the work done by friction in 10 seconds is 343 J.
Calculating change in kinetic energy:
The change in kinetic energy can be calculated using the formula:
Change in kinetic energy (ΔKE) = Work done by applied force (W) - Work done by friction (Wf)
The work done by applied force can be calculated using the formula:
Work done by applied force (W) = Force (F) × Distance (d)
Substituting the values:
Work done by applied force (W) = 7 N × 175 m
Work done by applied force (W) = 1225 J
Now, substituting the values into the equation for change in kinetic energy:
Change in kinetic energy (ΔKE) = 1225 J - 343 J
Change in kinetic energy (ΔKE) = 882 J
Therefore, the change in kinetic energy in 10 seconds is 882 J.
Conclusion:
- Mass of the body (m) = 2 kg
- Applied force (F) = 7 N
- Coefficient of kinetic friction (μ) = 0.1
- Time duration (t) = 10 s
Calculating work done by friction:
The work done by friction can be calculated using the formula:
Work done by friction (Wf) = Force of friction (f) × Distance (d)
The force of friction can be determined using the formula:
Force of friction (f) = μ × Normal force (N)
The normal force can be calculated as the product of the mass and acceleration due to gravity:
Normal force (N) = m × g
Where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Now, substituting the values into the equations:
Normal force (N) = 2 kg × 9.8 m/s^2 = 19.6 N
Force of friction (f) = 0.1 × 19.6 N = 1.96 N
To calculate the distance, we can use the equation of motion:
Distance (d) = Initial velocity (u) × time (t) + (1/2) × acceleration (a) × time squared (t^2)
Since the body is initially at rest, the initial velocity (u) is 0 m/s.
The acceleration (a) can be calculated using Newton's second law:
Force (F) = mass (m) × acceleration (a)
Substituting the values:
7 N = 2 kg × a
a = 3.5 m/s^2
Now, substituting the values into the equation of motion:
Distance (d) = 0 m/s × 10 s + (1/2) × 3.5 m/s^2 × (10 s)^2
Distance (d) = 0 + (1/2) × 3.5 m/s^2 × 100 s^2
Distance (d) = 0 + 175 m
Distance (d) = 175 m
Now, substituting the values into the equation for work done by friction:
Work done by friction (Wf) = 1.96 N × 175 m
Work done by friction (Wf) = 343 J
Therefore, the work done by friction in 10 seconds is 343 J.
Calculating change in kinetic energy:
The change in kinetic energy can be calculated using the formula:
Change in kinetic energy (ΔKE) = Work done by applied force (W) - Work done by friction (Wf)
The work done by applied force can be calculated using the formula:
Work done by applied force (W) = Force (F) × Distance (d)
Substituting the values:
Work done by applied force (W) = 7 N × 175 m
Work done by applied force (W) = 1225 J
Now, substituting the values into the equation for change in kinetic energy:
Change in kinetic energy (ΔKE) = 1225 J - 343 J
Change in kinetic energy (ΔKE) = 882 J
Therefore, the change in kinetic energy in 10 seconds is 882 J.
Conclusion:
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A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7N on a table with coefficient of kinetic friction = 0.1. Work done by friction in 10s and change in K.E. of the body in 10s are respectivelya)246.9 J, 635 Jb)200 J, 600 Jc)295 J, 625 Jd)500 J, 600 JCorrect answer is option 'A'. Can you explain this answer?
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A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7N on a table with coefficient of kinetic friction = 0.1. Work done by friction in 10s and change in K.E. of the body in 10s are respectivelya)246.9 J, 635 Jb)200 J, 600 Jc)295 J, 625 Jd)500 J, 600 JCorrect answer is option 'A'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7N on a table with coefficient of kinetic friction = 0.1. Work done by friction in 10s and change in K.E. of the body in 10s are respectivelya)246.9 J, 635 Jb)200 J, 600 Jc)295 J, 625 Jd)500 J, 600 JCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7N on a table with coefficient of kinetic friction = 0.1. Work done by friction in 10s and change in K.E. of the body in 10s are respectivelya)246.9 J, 635 Jb)200 J, 600 Jc)295 J, 625 Jd)500 J, 600 JCorrect answer is option 'A'. Can you explain this answer?.
A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7N on a table with coefficient of kinetic friction = 0.1. Work done by friction in 10s and change in K.E. of the body in 10s are respectivelya)246.9 J, 635 Jb)200 J, 600 Jc)295 J, 625 Jd)500 J, 600 JCorrect answer is option 'A'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7N on a table with coefficient of kinetic friction = 0.1. Work done by friction in 10s and change in K.E. of the body in 10s are respectivelya)246.9 J, 635 Jb)200 J, 600 Jc)295 J, 625 Jd)500 J, 600 JCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7N on a table with coefficient of kinetic friction = 0.1. Work done by friction in 10s and change in K.E. of the body in 10s are respectivelya)246.9 J, 635 Jb)200 J, 600 Jc)295 J, 625 Jd)500 J, 600 JCorrect answer is option 'A'. Can you explain this answer?.
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