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An external force of 6N is applied on a body at rest at time t=0. At t=1s direction of force is reversed. Find the speed of body when it comes back to the initial position. (mass= 2kg, coefficient of kinetic friction= 0.2)?
a) 0
b) √2 m/s
c) √1.2 m/s
d) 0.5 m/s?
a) 0
b) √2 m/s
c) √1.2 m/s
d) 0.5 m/s?
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An external force of 6N is applied on a body at rest at time t=0. At t...
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An external force of 6N is applied on a body at rest at time t=0. At t...
Given:
- External force applied: 6N
- Time when force direction is reversed: t=1s
- Mass of the body: 2kg
- Coefficient of kinetic friction: 0.2
To find:
The speed of the body when it comes back to the initial position.
Approach:
1. Calculate the acceleration of the body when the force is applied in the initial direction.
2. Calculate the deceleration of the body when the force direction is reversed.
3. Calculate the time taken for the body to stop after the force direction is reversed.
4. Calculate the distance covered by the body during this time period.
5. Calculate the time taken for the body to come back to the initial position.
6. Calculate the distance covered by the body during this time period.
7. Calculate the speed of the body when it comes back to the initial position.
Solution:
1. Calculate the acceleration of the body when the force is applied in the initial direction:
- The external force applied: F = 6N
- Mass of the body: m = 2kg
- Acceleration using Newton's second law: a = F/m = 6/2 = 3 m/s²
2. Calculate the deceleration of the body when the force direction is reversed:
- The external force applied after t=1s in the opposite direction: F' = -6N
- Acceleration using Newton's second law: a' = F'/m = -6/2 = -3 m/s² (negative sign indicates opposite direction)
3. Calculate the time taken for the body to stop after the force direction is reversed:
- The initial velocity of the body when the force direction is reversed: u = 0 m/s (at rest)
- The final velocity of the body when it stops: v = ?
- Using the equation of motion: v = u + at, where a = a' and u = 0
- v = 0 + (-3)t
- v = -3t
- When the body comes to a stop, v = 0, so -3t = 0
- t = 0
4. Calculate the distance covered by the body during this time period:
- Using the equation of motion: s = ut + (1/2)at², where a = a' and u = 0
- s = 0 + (1/2)(-3)(0)²
- s = 0
5. Calculate the time taken for the body to come back to the initial position:
- The initial velocity of the body when it starts moving in the opposite direction: u = 0 m/s
- The final velocity of the body when it comes back to the initial position: v = ?
- Using the equation of motion: v = u + at, where a = a' and u = 0
- v = 0 + (-3)t
- v = -3t
- When the body comes back to the initial position, v = 0, so -3t = 0
- t = 0
6. Calculate the distance covered by the body during this time period:
- Using
- External force applied: 6N
- Time when force direction is reversed: t=1s
- Mass of the body: 2kg
- Coefficient of kinetic friction: 0.2
To find:
The speed of the body when it comes back to the initial position.
Approach:
1. Calculate the acceleration of the body when the force is applied in the initial direction.
2. Calculate the deceleration of the body when the force direction is reversed.
3. Calculate the time taken for the body to stop after the force direction is reversed.
4. Calculate the distance covered by the body during this time period.
5. Calculate the time taken for the body to come back to the initial position.
6. Calculate the distance covered by the body during this time period.
7. Calculate the speed of the body when it comes back to the initial position.
Solution:
1. Calculate the acceleration of the body when the force is applied in the initial direction:
- The external force applied: F = 6N
- Mass of the body: m = 2kg
- Acceleration using Newton's second law: a = F/m = 6/2 = 3 m/s²
2. Calculate the deceleration of the body when the force direction is reversed:
- The external force applied after t=1s in the opposite direction: F' = -6N
- Acceleration using Newton's second law: a' = F'/m = -6/2 = -3 m/s² (negative sign indicates opposite direction)
3. Calculate the time taken for the body to stop after the force direction is reversed:
- The initial velocity of the body when the force direction is reversed: u = 0 m/s (at rest)
- The final velocity of the body when it stops: v = ?
- Using the equation of motion: v = u + at, where a = a' and u = 0
- v = 0 + (-3)t
- v = -3t
- When the body comes to a stop, v = 0, so -3t = 0
- t = 0
4. Calculate the distance covered by the body during this time period:
- Using the equation of motion: s = ut + (1/2)at², where a = a' and u = 0
- s = 0 + (1/2)(-3)(0)²
- s = 0
5. Calculate the time taken for the body to come back to the initial position:
- The initial velocity of the body when it starts moving in the opposite direction: u = 0 m/s
- The final velocity of the body when it comes back to the initial position: v = ?
- Using the equation of motion: v = u + at, where a = a' and u = 0
- v = 0 + (-3)t
- v = -3t
- When the body comes back to the initial position, v = 0, so -3t = 0
- t = 0
6. Calculate the distance covered by the body during this time period:
- Using
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An external force of 6N is applied on a body at rest at time t=0. At t=1s direction of force is reversed. Find the speed of body when it comes back to the initial position. (mass= 2kg, coefficient of kinetic friction= 0.2)?a) 0b) √2 m/sc) √1.2 m/sd) 0.5 m/s?
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An external force of 6N is applied on a body at rest at time t=0. At t=1s direction of force is reversed. Find the speed of body when it comes back to the initial position. (mass= 2kg, coefficient of kinetic friction= 0.2)?a) 0b) √2 m/sc) √1.2 m/sd) 0.5 m/s? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about An external force of 6N is applied on a body at rest at time t=0. At t=1s direction of force is reversed. Find the speed of body when it comes back to the initial position. (mass= 2kg, coefficient of kinetic friction= 0.2)?a) 0b) √2 m/sc) √1.2 m/sd) 0.5 m/s? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An external force of 6N is applied on a body at rest at time t=0. At t=1s direction of force is reversed. Find the speed of body when it comes back to the initial position. (mass= 2kg, coefficient of kinetic friction= 0.2)?a) 0b) √2 m/sc) √1.2 m/sd) 0.5 m/s?.
An external force of 6N is applied on a body at rest at time t=0. At t=1s direction of force is reversed. Find the speed of body when it comes back to the initial position. (mass= 2kg, coefficient of kinetic friction= 0.2)?a) 0b) √2 m/sc) √1.2 m/sd) 0.5 m/s? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about An external force of 6N is applied on a body at rest at time t=0. At t=1s direction of force is reversed. Find the speed of body when it comes back to the initial position. (mass= 2kg, coefficient of kinetic friction= 0.2)?a) 0b) √2 m/sc) √1.2 m/sd) 0.5 m/s? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An external force of 6N is applied on a body at rest at time t=0. At t=1s direction of force is reversed. Find the speed of body when it comes back to the initial position. (mass= 2kg, coefficient of kinetic friction= 0.2)?a) 0b) √2 m/sc) √1.2 m/sd) 0.5 m/s?.
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