A block of mass 2 kg slides down an inclined plane of inclination 60 degree find the work done by the force of gravity as the block comes down vertically by height of 1 metre
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Answer:
Given,
m=2kg
θ=30
0
μ=0.5
g=10m/s
2
The contact force is F
c
is the resultant of frictional force and normal force.
From the free body diagram,
Normal force, N=mgcosθ
N=2×10×cos30
0
N=17.320N
The friction force, F
r
=μmgcosθ
F
r
=0.5×2××10×cos30
0
F
r
=8.660N
N and F
r
are perpendicular to each other.
The contact force is, F
c
=
N
2
+F
r
2
F
c
=
(17.320)
2
+(8.699)
2
F
c
=19.381N
We can say approximately 20 N
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