A force F of magnitude 20N is applied to a block of mass 2Kg that lies on a rough horizontal surface as shown below in the figure. the coefficient of the kinetic friction between the block and surface is 0.4
a) what is the normal reaction force
b) what is the frictional force between the block and the surface
c) what is the magnitude of the acceleration
( g=10m/s^2 sin53=0.8 cos53=0.6 )
please i need it fast please....
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Answer: μ
k
=0.3
F=2T⇒T=10N
F
max
=MN
=0.3(20)
=6N
(T−f)=ma
(10−6)=2a
a=2m/s
2
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Answer:
A body of mass 2kg is resting on a rough horizontal surface. A force of 20N is now applied to it for 10s, parallel to the surface. If the coefficient of kinetic friction between the surfaces of contact is 0.2. Calculate <br> change in kinetic energy of the object in 10s.
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