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Two blocks of masses 4 kg and 2kg are connected by a heavy string of mass 3 kg and placed on a rough horizontal surface the 2 kg block is pulled with a constant force as shown in figure the coefficient of friction between the blocks at the ground is 0.5 what is the value of f so that the tension in the string is constant throughout during the motion of the blocks?
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Two blocks of masses 4 kg and 2kg are connected by a heavy string of m...
The frictional force acting on block of mass 2 kg = μmg=0.5x2x10=10 μmg=0.5x2x10=10 N
The force with which block of mass 4 kg is pulled is equal to ( F - 10 )
The frictional force acting on block of mass 4 kg = μmg=0.5x4x10=20μmg=0.5x4x10=20 N = ( F - 10 )
Thus, 
20=F−10⇒F=30 N
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Two blocks of masses 4 kg and 2kg are connected by a heavy string of m...
Tension in the string is constant throughout the motion of the blocks when the force applied to the 2 kg block is equal to the force of friction between the blocks and the ground. To find the value of force (f) that satisfies this condition, we need to consider the forces acting on each block and the friction between them.

Block 1 (4 kg):
- Weight (W1) = mass x acceleration due to gravity = 4 kg x 9.8 m/s^2 = 39.2 N
- Normal force (N1) = W1 (since the block is on a horizontal surface)
- Friction force (f1) = coefficient of friction (μ) x N1 = 0.5 x N1

Block 2 (2 kg):
- Weight (W2) = mass x acceleration due to gravity = 2 kg x 9.8 m/s^2 = 19.6 N
- Normal force (N2) = W2 (since the block is on a horizontal surface)
- Friction force (f2) = coefficient of friction (μ) x N2 = 0.5 x N2

Tension in the string (T) is equal to the force applied to the 2 kg block (f) since the string is assumed to be massless and inextensible.

To ensure constant tension, we need to equate the force applied to the 2 kg block (f) to the sum of the friction forces between the blocks and the ground (f1 and f2).

f = f1 + f2

Substituting the expressions for f1 and f2:

f = 0.5 x N1 + 0.5 x N2

Since N1 = W1 and N2 = W2, we can simplify the equation:

f = 0.5 x W1 + 0.5 x W2

f = 0.5 x (W1 + W2)

f = 0.5 x (39.2 N + 19.6 N)

f = 0.5 x 58.8 N

f = 29.4 N

Therefore, the value of force (f) that ensures constant tension in the string throughout the motion of the blocks is 29.4 N.
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Two blocks of masses 4 kg and 2kg are connected by a heavy string of mass 3 kg and placed on a rough horizontal surface the 2 kg block is pulled with a constant force as shown in figure the coefficient of friction between the blocks at the ground is 0.5 what is the value of f so that the tension in the string is constant throughout during the motion of the blocks?
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Two blocks of masses 4 kg and 2kg are connected by a heavy string of mass 3 kg and placed on a rough horizontal surface the 2 kg block is pulled with a constant force as shown in figure the coefficient of friction between the blocks at the ground is 0.5 what is the value of f so that the tension in the string is constant throughout during the motion of the blocks? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Two blocks of masses 4 kg and 2kg are connected by a heavy string of mass 3 kg and placed on a rough horizontal surface the 2 kg block is pulled with a constant force as shown in figure the coefficient of friction between the blocks at the ground is 0.5 what is the value of f so that the tension in the string is constant throughout during the motion of the blocks? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two blocks of masses 4 kg and 2kg are connected by a heavy string of mass 3 kg and placed on a rough horizontal surface the 2 kg block is pulled with a constant force as shown in figure the coefficient of friction between the blocks at the ground is 0.5 what is the value of f so that the tension in the string is constant throughout during the motion of the blocks?.
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