Question

Two rectangular blocks of masses \(40 \, kg\) and \(60 \, kg\) are connected by a string and kept on a frictionless horizontal table. If a force of \(1000 \, N\) is applied on the \(60 \, kg\) block away from the \(40 \, kg\) block, then the tension in the string is

• A. \(450 \, N\)
• B. \(400 \, N\)
• C. \(350 \, N\)
• D. \(500 \, N\)

Hint:

For connected blocks on a frictionless surface, first find the common acceleration using the total mass, then calculate tension using the block which is pulled only by the string.

Answer 1

The Correct Option is B

Step 1: See the two blocks as one system.
The string keeps both blocks moving together with one common acceleration, so to find that acceleration we treat the $40$ kg and $60$ kg blocks as a single mass being pushed by the $1000$ N force on a frictionless table.
Step 2: Find the total mass.
Total mass $=40+60=100$ kg.
Step 3: Find the common acceleration.
By Newton's second law, $a=\dfrac{F}{m}=\dfrac{1000}{100}=10$ m/s$^2$.
Step 4: Isolate the trailing block.
The $40$ kg block is dragged only by the string tension $T$, since the applied force acts on the $60$ kg block. The string is the only horizontal force on the $40$ kg block.
Step 5: Apply Newton's second law to it.
\[ T=40\times 10=400\ \text{N} \]
Step 6: Conclude.
The tension needed to accelerate the $40$ kg block at $10$ m/s$^2$ is $400$ N.
\[ \boxed{400\ \text{N}} \]