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A 2000kg car is driving 1000m along a horizontal road at 25m/s. If the road friction exert a 1000N force on the car how much work is done and how much power is required? what is the friction coefficient between the road and the tyre.( disregard wind resistance)?
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A 2000kg car is driving 1000m along a horizontal road at 25m/s. If the...
**Work Done by Friction Force**
To calculate the work done by the friction force, we need to use the formula:
Work = Force x Distance
Given:
Mass of the car (m) = 2000 kg
Distance traveled (d) = 1000 m
Speed of the car (v) = 25 m/s
Force of friction (F) = 1000 N
Using the formula for work, we can substitute the given values:
Work = 1000 N x 1000 m
Work = 1,000,000 J
Therefore, the work done by the friction force is 1,000,000 Joules.
**Power Required**
To calculate the power required, we need to use the formula:
Power = Work / Time
Given:
Work done (W) = 1,000,000 J
Time taken (t) = distance / speed = 1000 m / 25 m/s = 40 s
Using the formula for power, we can substitute the given values:
Power = 1,000,000 J / 40 s
Power = 25,000 W
Therefore, the power required is 25,000 Watts.
**Friction Coefficient**
To find the friction coefficient between the road and the tire, we need to use the formula:
Force of Friction = Friction Coefficient x Normal Force
Given:
Force of friction (F) = 1000 N
Mass of the car (m) = 2000 kg
Acceleration due to gravity (g) = 9.8 m/s²
The weight of the car is given by:
Weight (W) = mass x acceleration due to gravity
W = 2000 kg x 9.8 m/s²
W = 19600 N
Since the normal force is equal to the weight of the car, the normal force is 19600 N.
Using the formula for the force of friction, we can substitute the given values:
1000 N = Friction Coefficient x 19600 N
Solving for the friction coefficient:
Friction Coefficient = 1000 N / 19600 N
Friction Coefficient ≈ 0.051
Therefore, the friction coefficient between the road and the tire is approximately 0.051.
To calculate the work done by the friction force, we need to use the formula:
Work = Force x Distance
Given:
Mass of the car (m) = 2000 kg
Distance traveled (d) = 1000 m
Speed of the car (v) = 25 m/s
Force of friction (F) = 1000 N
Using the formula for work, we can substitute the given values:
Work = 1000 N x 1000 m
Work = 1,000,000 J
Therefore, the work done by the friction force is 1,000,000 Joules.
**Power Required**
To calculate the power required, we need to use the formula:
Power = Work / Time
Given:
Work done (W) = 1,000,000 J
Time taken (t) = distance / speed = 1000 m / 25 m/s = 40 s
Using the formula for power, we can substitute the given values:
Power = 1,000,000 J / 40 s
Power = 25,000 W
Therefore, the power required is 25,000 Watts.
**Friction Coefficient**
To find the friction coefficient between the road and the tire, we need to use the formula:
Force of Friction = Friction Coefficient x Normal Force
Given:
Force of friction (F) = 1000 N
Mass of the car (m) = 2000 kg
Acceleration due to gravity (g) = 9.8 m/s²
The weight of the car is given by:
Weight (W) = mass x acceleration due to gravity
W = 2000 kg x 9.8 m/s²
W = 19600 N
Since the normal force is equal to the weight of the car, the normal force is 19600 N.
Using the formula for the force of friction, we can substitute the given values:
1000 N = Friction Coefficient x 19600 N
Solving for the friction coefficient:
Friction Coefficient = 1000 N / 19600 N
Friction Coefficient ≈ 0.051
Therefore, the friction coefficient between the road and the tire is approximately 0.051.
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A 2000kg car is driving 1000m along a horizontal road at 25m/s. If the road friction exert a 1000N force on the car how much work is done and how much power is required? what is the friction coefficient between the road and the tyre.( disregard wind resistance)?
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A 2000kg car is driving 1000m along a horizontal road at 25m/s. If the road friction exert a 1000N force on the car how much work is done and how much power is required? what is the friction coefficient between the road and the tyre.( disregard wind resistance)? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A 2000kg car is driving 1000m along a horizontal road at 25m/s. If the road friction exert a 1000N force on the car how much work is done and how much power is required? what is the friction coefficient between the road and the tyre.( disregard wind resistance)? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 2000kg car is driving 1000m along a horizontal road at 25m/s. If the road friction exert a 1000N force on the car how much work is done and how much power is required? what is the friction coefficient between the road and the tyre.( disregard wind resistance)?.
A 2000kg car is driving 1000m along a horizontal road at 25m/s. If the road friction exert a 1000N force on the car how much work is done and how much power is required? what is the friction coefficient between the road and the tyre.( disregard wind resistance)? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A 2000kg car is driving 1000m along a horizontal road at 25m/s. If the road friction exert a 1000N force on the car how much work is done and how much power is required? what is the friction coefficient between the road and the tyre.( disregard wind resistance)? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 2000kg car is driving 1000m along a horizontal road at 25m/s. If the road friction exert a 1000N force on the car how much work is done and how much power is required? what is the friction coefficient between the road and the tyre.( disregard wind resistance)?.
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