Question

In the following case, find magnitude of friction on the block. (G
F=20N
370
10kg
=0.5
4 =0.4​

Answer 1

Answer:

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

i think this is the right answer