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A block of mass 2 kg is placed on the fl...

A block of mass 2 kg is placed on the floor. The coefficient of static friction is 0.4. A force F of 2.5 N is applied on the block as shown in Fig. 15.4.4. The force of friction between the block and the floor is `(g = 9.8 m//s^(2))`

A

2.8 N

B

8 N

C

2N

D

zero

Text Solution

Verified by Experts

The correct Answer is:
A
(a) `f_("max")=mu mg=0.4xx2xx10=8N`
Since the applied force is less then `f_("max")`, force of friction will be equal to the applied force or 2.8 N.
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DC PANDEY ENGLISH-LAWS OF MOTION-Check point 5.4
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