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A motorcycle of mass of 2000 kg is moving over a horizontal road,with uniform velocity. If this motorcycle has to be stopped with a negative acceleration of 1.5 m/s square,then what is the force of friction between the tires and the road?
Most Upvoted Answer
A motorcycle of mass of 2000 kg is moving over a horizontal road,with ...
Solution:
Given,
Mass of motorcycle, m = 2000 kg
Negative acceleration, a = -1.5 m/s^2
To find,
Force of friction, f
Using Newton's second law of motion, we know that the force required to stop the motorcycle is given by:
Force = mass x acceleration
Therefore, the force required to stop the motorcycle is:
Force = m x a
Force = 2000 kg x (-1.5 m/s^2)
Force = -3000 N
Since the force required is negative, it means that the force of friction acting on the motorcycle is in the opposite direction to its motion.
Therefore, the force of friction between the tires and the road is:
f = -3000 N
Explanation:
Newton's second law of motion states that the force acting on an object is directly proportional to its mass and acceleration. In this problem, the mass of the motorcycle is given as 2000 kg and the negative acceleration required to stop the motorcycle is given as 1.5 m/s^2. Using this information, we can calculate the force required to stop the motorcycle using the formula F = ma.
The force required to stop the motorcycle is negative, which means that the force of friction acting on the motorcycle is in the opposite direction to its motion. Therefore, the force of friction between the tires and the road is equal in magnitude but opposite in direction to the force required to stop the motorcycle.
In summary, the force of friction between the tires and the road can be calculated using Newton's second law of motion, and in this problem, it is equal to -3000 N.
Given,
Mass of motorcycle, m = 2000 kg
Negative acceleration, a = -1.5 m/s^2
To find,
Force of friction, f
Using Newton's second law of motion, we know that the force required to stop the motorcycle is given by:
Force = mass x acceleration
Therefore, the force required to stop the motorcycle is:
Force = m x a
Force = 2000 kg x (-1.5 m/s^2)
Force = -3000 N
Since the force required is negative, it means that the force of friction acting on the motorcycle is in the opposite direction to its motion.
Therefore, the force of friction between the tires and the road is:
f = -3000 N
Explanation:
Newton's second law of motion states that the force acting on an object is directly proportional to its mass and acceleration. In this problem, the mass of the motorcycle is given as 2000 kg and the negative acceleration required to stop the motorcycle is given as 1.5 m/s^2. Using this information, we can calculate the force required to stop the motorcycle using the formula F = ma.
The force required to stop the motorcycle is negative, which means that the force of friction acting on the motorcycle is in the opposite direction to its motion. Therefore, the force of friction between the tires and the road is equal in magnitude but opposite in direction to the force required to stop the motorcycle.
In summary, the force of friction between the tires and the road can be calculated using Newton's second law of motion, and in this problem, it is equal to -3000 N.
Community Answer
A motorcycle of mass of 2000 kg is moving over a horizontal road,with ...
Acceleration, a = -1.5 m/s²
mass, m = 2000 kg
Force = ma
= 2000×(-1.5)
= -3000N
So force is 3000N acting in the opposite direction to the motion.
(negative sign shows that it is acting opposite to the motion of object)....
mass, m = 2000 kg
Force = ma
= 2000×(-1.5)
= -3000N
So force is 3000N acting in the opposite direction to the motion.
(negative sign shows that it is acting opposite to the motion of object)....
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A motorcycle of mass of 2000 kg is moving over a horizontal road,with uniform velocity. If this motorcycle has to be stopped with a negative acceleration of 1.5 m/s square,then what is the force of friction between the tires and the road?
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A motorcycle of mass of 2000 kg is moving over a horizontal road,with uniform velocity. If this motorcycle has to be stopped with a negative acceleration of 1.5 m/s square,then what is the force of friction between the tires and the road? for Class 9 2024 is part of Class 9 preparation. The Question and answers have been prepared according to the Class 9 exam syllabus. Information about A motorcycle of mass of 2000 kg is moving over a horizontal road,with uniform velocity. If this motorcycle has to be stopped with a negative acceleration of 1.5 m/s square,then what is the force of friction between the tires and the road? covers all topics & solutions for Class 9 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A motorcycle of mass of 2000 kg is moving over a horizontal road,with uniform velocity. If this motorcycle has to be stopped with a negative acceleration of 1.5 m/s square,then what is the force of friction between the tires and the road?.
A motorcycle of mass of 2000 kg is moving over a horizontal road,with uniform velocity. If this motorcycle has to be stopped with a negative acceleration of 1.5 m/s square,then what is the force of friction between the tires and the road? for Class 9 2024 is part of Class 9 preparation. The Question and answers have been prepared according to the Class 9 exam syllabus. Information about A motorcycle of mass of 2000 kg is moving over a horizontal road,with uniform velocity. If this motorcycle has to be stopped with a negative acceleration of 1.5 m/s square,then what is the force of friction between the tires and the road? covers all topics & solutions for Class 9 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A motorcycle of mass of 2000 kg is moving over a horizontal road,with uniform velocity. If this motorcycle has to be stopped with a negative acceleration of 1.5 m/s square,then what is the force of friction between the tires and the road?.
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