Physics Q86 – AP-EAMCET 2026

Home Physics Laws of MotionQuestion #207

Two blocks of masses

m

and

M

are connected by an inextensible light string. A constant horizontal force

f

acts on the block of mass

M

, then tension in the string is (Neglect friction)

Clue 1 of 3: System Acceleration

First, consider the two blocks as a single system to determine the common acceleration of both blocks under the influence of the external force.

🧠 Full Solution Path

Step 1: Calculate the System's Acceleration✦ Active

Since the two blocks are connected by an inextensible string and friction is neglected, they move together as a single system. The total mass of the system is the sum of the individual masses, $M_{t o t a l} = M + m$ . The constant horizontal force $f$ acts on the block of mass $M$ , making it the net external force on the system in the horizontal direction.

Using Newton's Second Law ( $F_{n e t} = M_{t o t a l} a$ ) for the entire system:

f = (M + m) a

Therefore, the acceleration of the system is:

a = \frac{f}{M + m}

💡 Teacher's Secret Hint

Remember that for connected bodies moving together, the acceleration of each body is the same as the acceleration of the system.

Step 2: Analyze Forces on Mass

m

○ Expand

Now, consider the free-body diagram for the block of mass $m$ . The forces acting on $m$ are its weight $m g$ downwards, the normal force $N_{m}$ upwards, and the tension $T$ in the string pulling it at an angle $θ$ with the horizontal.

We are interested in the horizontal motion, so we consider the horizontal component of the tension. The horizontal component of the tension $T$ is $T \cos θ$ .

💡 Teacher's Secret Hint

Always resolve forces into components along the direction of motion (or perpendicular to it) when dealing with inclined forces.

Step 3: Apply Newton's Second Law to Mass

m

○ Expand

Applying Newton's Second Law ( $F_{n e t, x} = m a_{x}$ ) to the horizontal motion of block $m$ :

T \cos θ = m a

Substitute the expression for acceleration $a$ from Step 1 into this equation:

T \cos θ = m (\frac{f}{M + m})

💡 Teacher's Secret Hint

Ensure that the acceleration $a$ used here is the common acceleration of the system, as derived in Step 1.

Step 4: Solve for Tension

T

○ Expand

Rearrange the equation from Step 3 to solve for the tension $T$ :

T = \frac{m f}{(M + m) \cos θ}

💡 Teacher's Secret Hint

Check your final expression for dimensional consistency. Tension is a force, so it should have units of force. This can help catch algebraic errors.

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Physics Question 86 – AP-EAMCET 2026

Clue 1 of 3: System Acceleration

🧠 Full Solution Path