1.

What will be the area of contact of a block weighing 20N, if it can apply a pressure of 2000Pa?

Answer»

Answer:

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Explanation:

a. F

fric

s

W=0.4∗20=8N

b. SINCE F

applied

fric

F

fric

=5N

c. MINIMUM force required to start the MOTION F=F

staticfric

=8N

d. Minimum force required to start the motion

F=F

kineticfric

k

W=0.2∗20=4N

e. F>F

staticfric

motion started.

F>F

kineticfric

hence motion continues.

F

fric

=4N



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