Free Body Diagrams
Mediumphysics
Two blocks of masses 4 kg (top) and 6 kg (bottom) are in contact on a frictionless surface. A horizontal 30 N force is applied to the 6 kg block. What is the contact force between the blocks?
Select the correct option:
Solution
Incorrect! Answer:
12 N
- System Acceleration (a): Treat the two blocks as a single system of mass M=4+6=10 kg.
- a=MFext=1030=3 m/s2.
- Isolate one block: To find the contact force (N), look at the 4 kg block (the one not directly pushed by 30N).
- FBD of 4 kg Block: The only horizontal force on it is the contact force from the 6 kg block.
- Fnet=m2a
- N=4 kg×3 m/s2=12 N.
- Verification: Net force on 6 kg block =30−12=18 N. Acceleration check: 18/6=3 m/s2 (Matches).
🔒 Solution Hidden from View
Submit your answer to unlock the detailed step-by-step solution.
About This Question
- Subject
- physics
- Chapter
- laws of motion
- Topic
- free body diagrams
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
12 N
- System Acceleration (a): Treat the two blocks as a single system of mass M=4+6=10 kg.
- a=MFext=1030=3 m/s2.
- Isolate one block: To find the contact force (N), look at the 4 kg block (the one not directly pushed by 30N).
- FBD of 4 kg Block: The only horizontal force on it is the contact force from the 6 kg block.
- Fnet=m2a
- N=4 kg×3 m/s2=12 N.
- Verification: Net force on 6 kg block =30−12=18 N. Acceleration check: 18/6=3 m/s2 (Matches).
This medium difficulty physics question is from the chapter laws of motion, covering the topic of free body diagrams. It appeared in the 2025 exam.
Looking for more practice? Explore all physics questions or browse laws of motion questions on RankGuru.