A force of 0.03 Newtown in a direction parallel to the line of greatest slope is applied to the block so that it moves up the plane. When the block has travelled a distance of 110 cm from its initial position the applied force is removed .the block moves on and comes to the rest again after travelling a further 25 cm
Calculate the work done by applied force
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Answer:
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:
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