The inclined plane with base angle 30 degree is used to lify a load of 500N usng 400N.What is the efficiency of the plane?
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Direction of electric field is not given exactly but it is given that it is in horizontal plane. Two scenarios are possible as shown in the attached diagram.
Scenario (A)
Net force perpendicular to the incline is zero as the block has no displacement perpendicular to the inclined plane.
From the free body diagram,
N+qEsinθ=mgcosθ
N=1×10×cos(30
o
)−0.01×100×sin(30
o
)
N=5
3
−0.5 N
Frictional force (sliding) acting is:
F
r
=μN
F
r
=0.2×(5
3
−0.5)
F
r
=
3
−0.01 N
From the free body diagram, net force down the incline is:
F
d
=mgsinθ+qEcosθ−F
r
F
d
=1×10×sin(30
o
)+0.01×100×cos(30
o
)−(
3
−0.01)
F
d
=5.01−
2
3
≈4.134 N
From newton's second law of motion, acceleration is:
a=
m
F
d
=
1
4.134
=4.134 m/s
2
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