A block of mass 2 kg is placed on the floor the coefficient of static friction is 0.4 if the force of 2.8 is applied on the block parallel force the force of the friction between the block and the floor is
shatrughansharma1968:
The block does not move at all then why we take friction equals to 2.8N
u hv unclear concepts. As ext force of 2.8 N is applied on body , body tends to move. It is friction who keeps it at rest and has magnitude of 2.8 N . Hope u get it
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concept of sliding friction
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Hey dear,
◆ Answer-
Frictional force = 2.8 N
◆ Explaination-
# Given-
m = 2 kg
μ = 0.4
Fext = 2.8 N
# Solution-
Limiting force of friction is calculated by-
Fs = μN = μmg
Fs = 0.4 × 2 × 9.8
Fs = 7.84 N
As two frictional forces can act on body i.e. Fs = 7.84 N and Fext = 2.8 N, the lower one (2.8N) will act actually.
Therefore, force of friction between block and floor is 2.8 N .
Hope this helps...
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