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A bag is gently dropped on a conveyor belt moving at a speed of 2 m/s. The coefficient of friction between the conveyor belt and bag is 0.4. Initially, the bag slips on the belt before it stops due to friction. The distance travelled by the bag on the belt during slipping motion is:
[Take g = 10 ms-2]
[Take g = 10 ms-2]
- a)2 m
- b)0.5 m
- c)3.2 m
- d)0.8 ms
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
A bag is gently dropped on a conveyor belt moving at a speed of 2 m/s....
To solve this problem, we can use the concept of friction and the equations of motion. Let's break down the problem into the following steps:
Step 1: Determine the acceleration of the bag
Since the bag is gently dropped on the conveyor belt, its initial velocity is zero. The bag comes to rest due to the friction between the belt and the bag. We can use the equation of motion to find the acceleration:
v^2 = u^2 + 2as
where
v = final velocity (0 m/s)
u = initial velocity (0 m/s)
a = acceleration
s = distance travelled
Rearranging the equation, we get:
a = -u^2 / (2s)
Since the bag is slipping, the frictional force opposes the motion and acts in the opposite direction to the velocity. Therefore, the acceleration is negative.
Step 2: Calculate the frictional force
The frictional force can be calculated using the equation:
f = μN
where
f = frictional force
μ = coefficient of friction (0.4)
N = normal force
The normal force is equal to the weight of the bag, which can be calculated as:
N = mg
where
m = mass of the bag
g = acceleration due to gravity (10 m/s^2)
Step 3: Equate the frictional force to the force causing the acceleration
The frictional force is equal to the mass of the bag multiplied by the acceleration:
f = ma
Substituting the values we have:
μN = ma
μmg = ma
Since the mass cancels out, we get:
μg = a
Step 4: Calculate the distance travelled during slipping motion
We can substitute the value of acceleration into the equation obtained in Step 1:
a = -u^2 / (2s)
μg = -u^2 / (2s)
Simplifying the equation, we get:
s = -u^2 / (2μg)
Substituting the given values:
s = -(0^2) / (2 * 0.4 * 10)
s = 0 / 8
s = 0.5 m
Therefore, the distance travelled by the bag on the belt during slipping motion is 0.5 m, which corresponds to option (B).
Step 1: Determine the acceleration of the bag
Since the bag is gently dropped on the conveyor belt, its initial velocity is zero. The bag comes to rest due to the friction between the belt and the bag. We can use the equation of motion to find the acceleration:
v^2 = u^2 + 2as
where
v = final velocity (0 m/s)
u = initial velocity (0 m/s)
a = acceleration
s = distance travelled
Rearranging the equation, we get:
a = -u^2 / (2s)
Since the bag is slipping, the frictional force opposes the motion and acts in the opposite direction to the velocity. Therefore, the acceleration is negative.
Step 2: Calculate the frictional force
The frictional force can be calculated using the equation:
f = μN
where
f = frictional force
μ = coefficient of friction (0.4)
N = normal force
The normal force is equal to the weight of the bag, which can be calculated as:
N = mg
where
m = mass of the bag
g = acceleration due to gravity (10 m/s^2)
Step 3: Equate the frictional force to the force causing the acceleration
The frictional force is equal to the mass of the bag multiplied by the acceleration:
f = ma
Substituting the values we have:
μN = ma
μmg = ma
Since the mass cancels out, we get:
μg = a
Step 4: Calculate the distance travelled during slipping motion
We can substitute the value of acceleration into the equation obtained in Step 1:
a = -u^2 / (2s)
μg = -u^2 / (2s)
Simplifying the equation, we get:
s = -u^2 / (2μg)
Substituting the given values:
s = -(0^2) / (2 * 0.4 * 10)
s = 0 / 8
s = 0.5 m
Therefore, the distance travelled by the bag on the belt during slipping motion is 0.5 m, which corresponds to option (B).
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Community Answer
A bag is gently dropped on a conveyor belt moving at a speed of 2 m/s....
μ = 0.4
Velocity of the conveyor belt = 2 m/s
Initially, when the bag is dropped on the conveyor belt, it starts slipping.
S
Motion with respect to belt:
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A bag is gently dropped on a conveyor belt moving at a speed of 2 m/s. The coefficient of friction between the conveyor belt and bag is 0.4. Initially, the bag slips on the belt before it stops due to friction. The distance travelled by the bag on the belt during slipping motion is:[Take g = 10 ms-2]a)2 mb)0.5 mc)3.2 md)0.8 msCorrect answer is option 'B'. Can you explain this answer?
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A bag is gently dropped on a conveyor belt moving at a speed of 2 m/s. The coefficient of friction between the conveyor belt and bag is 0.4. Initially, the bag slips on the belt before it stops due to friction. The distance travelled by the bag on the belt during slipping motion is:[Take g = 10 ms-2]a)2 mb)0.5 mc)3.2 md)0.8 msCorrect answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A bag is gently dropped on a conveyor belt moving at a speed of 2 m/s. The coefficient of friction between the conveyor belt and bag is 0.4. Initially, the bag slips on the belt before it stops due to friction. The distance travelled by the bag on the belt during slipping motion is:[Take g = 10 ms-2]a)2 mb)0.5 mc)3.2 md)0.8 msCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A bag is gently dropped on a conveyor belt moving at a speed of 2 m/s. The coefficient of friction between the conveyor belt and bag is 0.4. Initially, the bag slips on the belt before it stops due to friction. The distance travelled by the bag on the belt during slipping motion is:[Take g = 10 ms-2]a)2 mb)0.5 mc)3.2 md)0.8 msCorrect answer is option 'B'. Can you explain this answer?.
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A bag is gently dropped on a conveyor belt moving at a speed of 2 m/s. The coefficient of friction between the conveyor belt and bag is 0.4. Initially, the bag slips on the belt before it stops due to friction. The distance travelled by the bag on the belt during slipping motion is:[Take g = 10 ms-2]a)2 mb)0.5 mc)3.2 md)0.8 msCorrect answer is option 'B'. Can you explain this answer?, a detailed solution for A bag is gently dropped on a conveyor belt moving at a speed of 2 m/s. The coefficient of friction between the conveyor belt and bag is 0.4. Initially, the bag slips on the belt before it stops due to friction. The distance travelled by the bag on the belt during slipping motion is:[Take g = 10 ms-2]a)2 mb)0.5 mc)3.2 md)0.8 msCorrect answer is option 'B'. Can you explain this answer? has been provided alongside types of A bag is gently dropped on a conveyor belt moving at a speed of 2 m/s. The coefficient of friction between the conveyor belt and bag is 0.4. Initially, the bag slips on the belt before it stops due to friction. The distance travelled by the bag on the belt during slipping motion is:[Take g = 10 ms-2]a)2 mb)0.5 mc)3.2 md)0.8 msCorrect answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
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