A 2 kg block is pushed against a vertical wall by applying a horizontal force of 50 N. The coefficient of static friction between the block and the wall is 0.5. A force F is also applied on the block vertically upward (as shown in figure). The maximum value of F applied, so that the block does not move upward, will be :
(Given : g = 10 ms\( - \)2)
A 2 kg block is pushed against a vertical wall by applying a horizontal force of 50 N. The coefficient of static friction between the block and the wall is 0.5. A force F is also applied on the block vertically upward (as shown in figure). The maximum value of F applied, so that the block does not move upward, will be :
(Given : g = 10 ms\( - \)2)
Show Hint
- 10 N
- 20 N
- 25 N
- 45 N
The Correct Option is D
Solution and Explanation
Step 2: Weight \( mg=2\times10=20 \) N downward. When F is increased, the block tends to move upward, so friction acts downward.
Step 3: At the limiting case: \( F=mg+f_{max}=20+25=45 \) N.
Step 4: The maximum F so the block does not move upward is 45 N.
Answer: (D) 45 N
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