a block of mass 5kg is placed on a horizontal table .a person pushes the block from top by exerting a downward force of 10N on it.find the force exerted by the table on the body(g=10ms^-2).
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Answer:
Given,
Coefficient of friction, μ=0.2
Mass of block, M=5kg
Friction force, F
r
=μ(mg+Fsin45)=0.2×(5×9.8+
2
20
)=12.63 N
Net force, F
net
=Fcos45−F
r
=
2
20
−12.63=1.5 N
Acceleration, a=
m
F
net
=
5
1.5
=0.3 ms
−2
Apply kinematic equation of motion
V=U+at=0+
5
1.5
×15=4.5 ms
−1
Final velocity is 4.5 ms
−1
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