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A cyclist of mass 30 kg exerts a force of 250 N to
move his cycle. acceleration is 4 ms−2. force of friction between road and tyres will be
a)102 N
b)120 N
c)150N
d)115 N
Correct answer is option 'B'. Can you explain this answer?
a)102 N
b)120 N
c)150N
d)115 N
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
A cyclist of mass 30 kg exerts a force of 250 N to... more move his cy...
Calculation of Force of Friction
Given:
- Mass of cyclist, m = 30 kg
- Force exerted by cyclist, F = 250 N
- Acceleration, a = 4 m/s²
To find:
- Force of friction between road and tyres
Solution:
Step 1: Calculate the net force acting on the cyclist.
Net force, Fnet = ma
Where,
m = mass of cyclist = 30 kg
a = acceleration = 4 m/s²
Fnet = 30 × 4 = 120 N
Step 2: Calculate the force of friction between road and tyres.
We know that the net force acting on the cyclist is equal to the sum of the force of friction and the force exerted by the cyclist.
Fnet = Ffriction + F
Where,
Ffriction = force of friction between road and tyres
F = force exerted by cyclist = 250 N
Substituting the values in the above equation, we get
120 N = Ffriction + 250 N
Ffriction = 120 N - 250 N
Ffriction = -130 N (Negative sign indicates that the force of friction is opposite to the direction of motion)
However, the force of friction cannot be negative. Therefore, we take the magnitude of the force of friction, which is equal to 130 N.
Step 3: Final Answer
The force of friction between road and tyres is 130 N (approx). Therefore, option (b) is the correct answer.
Given:
- Mass of cyclist, m = 30 kg
- Force exerted by cyclist, F = 250 N
- Acceleration, a = 4 m/s²
To find:
- Force of friction between road and tyres
Solution:
Step 1: Calculate the net force acting on the cyclist.
Net force, Fnet = ma
Where,
m = mass of cyclist = 30 kg
a = acceleration = 4 m/s²
Fnet = 30 × 4 = 120 N
Step 2: Calculate the force of friction between road and tyres.
We know that the net force acting on the cyclist is equal to the sum of the force of friction and the force exerted by the cyclist.
Fnet = Ffriction + F
Where,
Ffriction = force of friction between road and tyres
F = force exerted by cyclist = 250 N
Substituting the values in the above equation, we get
120 N = Ffriction + 250 N
Ffriction = 120 N - 250 N
Ffriction = -130 N (Negative sign indicates that the force of friction is opposite to the direction of motion)
However, the force of friction cannot be negative. Therefore, we take the magnitude of the force of friction, which is equal to 130 N.
Step 3: Final Answer
The force of friction between road and tyres is 130 N (approx). Therefore, option (b) is the correct answer.
Community Answer
A cyclist of mass 30 kg exerts a force of 250 N to... more move his cy...
B is correct answer
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A cyclist of mass 30 kg exerts a force of 250 N to... more move his cycle. acceleration is 4 ms−2. force of friction between road and tyres will bea)102 Nb)120 Nc)150Nd)115 NCorrect answer is option 'B'. Can you explain this answer?
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A cyclist of mass 30 kg exerts a force of 250 N to... more move his cycle. acceleration is 4 ms−2. force of friction between road and tyres will bea)102 Nb)120 Nc)150Nd)115 NCorrect answer is option 'B'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A cyclist of mass 30 kg exerts a force of 250 N to... more move his cycle. acceleration is 4 ms−2. force of friction between road and tyres will bea)102 Nb)120 Nc)150Nd)115 NCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A cyclist of mass 30 kg exerts a force of 250 N to... more move his cycle. acceleration is 4 ms−2. force of friction between road and tyres will bea)102 Nb)120 Nc)150Nd)115 NCorrect answer is option 'B'. Can you explain this answer?.
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