A straight wire of mass 200g and length 1.5m carries a current of 2A. It is suspended in mid air by a uniform horizontal magnetic field B.
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A rigid body of mass 0.3 kg is taken slowly up an inclined plane of length 10 m and height 5 m (assuming the applied force to be parallel to the inclined plane), and then allowed to slide down to the bottom again. The coefficient of friction between the body and the plane is 0.15. using g = 9.8 m/s
2
find the
(a) work done y the gravitational force over the round trip
(b) work done by the applied force over the upward journey
(c) work done by frictional force over the round trip
(d) kinetic energy of the body at the end of the trip?
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Hard
Solution
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It is given that
m=0.3Kg
h=10m
l=5m
μ=0.15
(a) As the displacement over the round trip is zero so, work done by the gravitational force over the round trip is zero.
(b) work done by the applied force over the upward journey
W
up
=F×d
=(mgsinθ+f)h
=mg(sinθ+μcosθ)h
=0.3×9.8×10((
10
5
)+0.15(
10
75
))
=29.4(0.5+0.129)
=18.51J
(c) work done by frictional force over the round trip
W=2fh
=2(μmghcosθ)
=2×0.15×0.3×9.8×
10
75
×10
=7.638J
(d) kinetic energy of the body at the end of the trip
K=
2
1
mv
2
From newton’s second law
mgsinθ−f=ma
mgsinθ−μmgcosθ=ma
a=g(sinθ−μcosθ)
a=9.8(0.5−0.15×
10
75
)
a=3.62ms
−2
v
2
−u
2
=2as
v
2
=72.4
v=8.5m/s
K=
2
1
×0.3×72.4K=10.86J
Explanation:
Only a best friend can see the pain behind a fake smile. Author Unknown. “A long sleeve a fake smile and everyone thinks you are fine
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