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The block C shown in figure is ascending with an acceleration a=3m/s² by means of some motor not shown here. The bodies of A and B have masses 10 kg and 5 kg respectively, assuming pulleys and strings are massless and friction is absent everywhere. Then find acceleration of body A (in m/s²)?
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The block C shown in figure is ascending with an acceleration a=3m/s² ...
Acceleration of Body A:
Given:
Mass of Body A (mA) = 10 kg
Mass of Body B (mB) = 5 kg
Acceleration of Block C (aC) = 3 m/s²
To find:
Acceleration of Body A (aA)
Analysis:
1. The system consists of three bodies: A, B, and C.
2. Body A is connected to Body B by a string passing over a pulley, and Body B is connected to Block C by another string passing over another pulley.
3. The pulleys and strings are assumed to be massless and frictionless.
4. The system is in equilibrium, meaning the total forces acting on each body are balanced.
Using Newton's second law of motion:
1. For Body A:
- The only force acting on Body A is the tension in the string.
- The tension in the string is equal to the weight of Body B, as the system is in equilibrium.
- Therefore, the net force on Body A is given by:
FnetA = Tension = Weight of Body B = mB * g
where g is the acceleration due to gravity (approximated as 9.8 m/s²).
- Using Newton's second law, we have:
FnetA = mA * aA
where aA is the acceleration of Body A.
2. For Body B:
- The net force on Body B is the difference between the tension in the string (pulling it upwards) and the weight of Body B (pulling it downwards).
- Therefore, the net force on Body B is given by:
FnetB = Tension - Weight of Body B = mA * aA - mB * g
- Using Newton's second law, we have:
FnetB = mB * aB
where aB is the acceleration of Body B.
3. For Block C:
- The net force on Block C is equal to the tension in the string.
- Therefore, the net force on Block C is given by:
FnetC = Tension = mC * aC
where mC is the mass of Block C and aC is its acceleration.
Equating the tensions in the strings:
Tension = Tension
mA * g = mB * aB
aB = (mA * g) / mB
Substituting the value of aB in the equation for FnetB:
mB * aB = mA * aA - mB * g
mB * [(mA * g) / mB] = mA * aA - mB * g
mA * g = mA * aA - mB * g
aA = (mA * g + mB * g) / mA
aA = g * (mA + mB) / mA
Substituting the values:
aA = 9.8 m/s² * (10 kg + 5 kg) / 10 kg
aA = 9.8 m/s² * 15 kg / 10 kg
aA = 14.7 m/s²
Therefore, the acceleration of Body A is 14.7 m/s².
Given:
Mass of Body A (mA) = 10 kg
Mass of Body B (mB) = 5 kg
Acceleration of Block C (aC) = 3 m/s²
To find:
Acceleration of Body A (aA)
Analysis:
1. The system consists of three bodies: A, B, and C.
2. Body A is connected to Body B by a string passing over a pulley, and Body B is connected to Block C by another string passing over another pulley.
3. The pulleys and strings are assumed to be massless and frictionless.
4. The system is in equilibrium, meaning the total forces acting on each body are balanced.
Using Newton's second law of motion:
1. For Body A:
- The only force acting on Body A is the tension in the string.
- The tension in the string is equal to the weight of Body B, as the system is in equilibrium.
- Therefore, the net force on Body A is given by:
FnetA = Tension = Weight of Body B = mB * g
where g is the acceleration due to gravity (approximated as 9.8 m/s²).
- Using Newton's second law, we have:
FnetA = mA * aA
where aA is the acceleration of Body A.
2. For Body B:
- The net force on Body B is the difference between the tension in the string (pulling it upwards) and the weight of Body B (pulling it downwards).
- Therefore, the net force on Body B is given by:
FnetB = Tension - Weight of Body B = mA * aA - mB * g
- Using Newton's second law, we have:
FnetB = mB * aB
where aB is the acceleration of Body B.
3. For Block C:
- The net force on Block C is equal to the tension in the string.
- Therefore, the net force on Block C is given by:
FnetC = Tension = mC * aC
where mC is the mass of Block C and aC is its acceleration.
Equating the tensions in the strings:
Tension = Tension
mA * g = mB * aB
aB = (mA * g) / mB
Substituting the value of aB in the equation for FnetB:
mB * aB = mA * aA - mB * g
mB * [(mA * g) / mB] = mA * aA - mB * g
mA * g = mA * aA - mB * g
aA = (mA * g + mB * g) / mA
aA = g * (mA + mB) / mA
Substituting the values:
aA = 9.8 m/s² * (10 kg + 5 kg) / 10 kg
aA = 9.8 m/s² * 15 kg / 10 kg
aA = 14.7 m/s²
Therefore, the acceleration of Body A is 14.7 m/s².
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The block C shown in figure is ascending with an acceleration a=3m/s² by means of some motor not shown here. The bodies of A and B have masses 10 kg and 5 kg respectively, assuming pulleys and strings are massless and friction is absent everywhere. Then find acceleration of body A (in m/s²)?
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The block C shown in figure is ascending with an acceleration a=3m/s² by means of some motor not shown here. The bodies of A and B have masses 10 kg and 5 kg respectively, assuming pulleys and strings are massless and friction is absent everywhere. Then find acceleration of body A (in m/s²)? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The block C shown in figure is ascending with an acceleration a=3m/s² by means of some motor not shown here. The bodies of A and B have masses 10 kg and 5 kg respectively, assuming pulleys and strings are massless and friction is absent everywhere. Then find acceleration of body A (in m/s²)? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The block C shown in figure is ascending with an acceleration a=3m/s² by means of some motor not shown here. The bodies of A and B have masses 10 kg and 5 kg respectively, assuming pulleys and strings are massless and friction is absent everywhere. Then find acceleration of body A (in m/s²)?.
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