Question

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

Answer 1

Answer:

Hello my friend

Explanation:

a. F

fric

=μ

s

W=0.4∗20=8N

b. Since F

applied

<F

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

=μ

k

W=0.2∗20=4N

e. F>F

staticfric

motion started.

F>F

kineticfric

hence motion continues.

F

fric

=4N

Answer 2

FORCE=20N

PRESSURE =2000pa

AREA=FORCE

PRESSURE

= 20N

2000pa

=0.01

.

. . AREA=0.01m²

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